Arithmetic of finance

1. Revision

ID is: 2815 Seed is: 3551

Finding the length of an investment

At some time in the past, Mike deposited $\text{N}\llap{=}\,\text{2,300}$ into a savings account at The Bank of Money Money. The account earns interest at a rate of $\text{6.42}\%$ compounded yearly. How long ago did Mike open the account if the balance is now $\text{N}\llap{=}\,\text{3,074.94}$ ? Give your answer in years and months.

Answer:

The account was opened years and month(s) ago.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

What kind of interest was earned? Which formula do we need for this calculation?

STEP: <no title>
[−0 points ⇒ 4 / 4 points left]

From the question statement, we know that this is a compound interest calculation. We read the question carefully and write down the given information:

\begin{aligned} A & = P (1 + i)^{n} \\ Where: \\ A & = 3,074.94 \\ P & = 2,300 \\ i & = 6.42 % = 0.0642 \\ n & = ? \end{aligned}

$\begin{aligned} A & = P(1+i)^n \\ \text{Where: } & \\ A & = \text{3,074.94} \\ P & = \text{2,300} \\ i & = \text{6.42}\% = \text{0.0642} \\ n & = \text{?} \end{aligned}$

STEP: <no title>
[−2 points ⇒ 2 / 4 points left]

Substitute the known values into the equation and solve for $n$ .

\begin{aligned} 3,074.94 & = 2,300 {(1 + 0.0642)}^{n} \\ 3,074.94 & = 2,300 (1.0642)^{n} \\ \frac{3,074.94}{2,300} & = (1.0642)^{n} \end{aligned}

$\begin{aligned} \text{3,074.94} & = \text{2,300} \left( 1 + \text{0.0642} \right) ^ {n} \\ \text{3,074.94} & = \text{2,300}(\text{1.0642})^{n} \\ \frac{\text{3,074.94}}{\text{2,300}} & = (\text{1.0642})^{n} \end{aligned}$

STEP: <no title>
[−1 point ⇒ 1 / 4 points left]

At this stage, we convert to logarithmic form so that $n$ is the subject of the formula:

\begin{aligned} n & = \log_{1.0642} (\frac{3,074.94}{2,300}) \\ n & = 4.66666... \end{aligned}

$\begin{aligned} n &= \log_{\text{1.0642}}{\left(\frac{\text{3,074.94}}{\text{2,300}}\right)} \\ n &= \text{4.66666...} \end{aligned}$

STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

We know that Mike left the money in the account for about $\text{4.67}$ years. However, we need to give the answer in terms of years and months.

$\text{4.66666...}$ years $\; = \;$ $\text{4}$ years and $\text{0.66666...}$ of a year (which will be in months).

We know that there are $\text{12}$ months in a year. To convert $\text{0.66666...}$ of a year into months, we do the following calculation:

\begin{matrix} 0.66666... of a year \times (\frac{12 months}{year}) = 8 month(s) \end{matrix}

$\begin{array}{c}\text{0.66666...}\text{ of a year} \times \left( \frac{12\text{ months}}{\text{year}} \right) = \text{8} \text{ month(s)}\end{array}$

Therefore, Mike deposited the money into the account $\text{4}$ years and $\text{8}$ month(s) ago.

Submit your answer as: and

ID is: 2815 Seed is: 476

Finding the length of an investment

At some time in the past, Wethu deposited $\text{N}\llap{=}\,\text{2,750}$ into a savings account at The Bank of Money Money. The account earns interest at a rate of $\text{7.81}\%$ compounded yearly. How long ago did Wethu open the account if the balance is now $\text{N}\llap{=}\,\text{4,211.15}$ ? Give your answer in years and months.

Answer:

The account was opened years and month(s) ago.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

What kind of interest was earned? Which formula do we need for this calculation?

STEP: <no title>
[−0 points ⇒ 4 / 4 points left]

From the question statement, we know that this is a compound interest calculation. We read the question carefully and write down the given information:

\begin{aligned} A & = P (1 + i)^{n} \\ Where: \\ A & = 4,211.15 \\ P & = 2,750 \\ i & = 7.81 % = 0.0781 \\ n & = ? \end{aligned}

$\begin{aligned} A & = P(1+i)^n \\ \text{Where: } & \\ A & = \text{4,211.15} \\ P & = \text{2,750} \\ i & = \text{7.81}\% = \text{0.0781} \\ n & = \text{?} \end{aligned}$

STEP: <no title>
[−2 points ⇒ 2 / 4 points left]

Substitute the known values into the equation and solve for $n$ .

\begin{aligned} 4,211.15 & = 2,750 {(1 + 0.0781)}^{n} \\ 4,211.15 & = 2,750 (1.0781)^{n} \\ \frac{4,211.15}{2,750} & = (1.0781)^{n} \end{aligned}

$\begin{aligned} \text{4,211.15} & = \text{2,750} \left( 1 + \text{0.0781} \right) ^ {n} \\ \text{4,211.15} & = \text{2,750}(\text{1.0781})^{n} \\ \frac{\text{4,211.15}}{\text{2,750}} & = (\text{1.0781})^{n} \end{aligned}$

STEP: <no title>
[−1 point ⇒ 1 / 4 points left]

At this stage, we convert to logarithmic form so that $n$ is the subject of the formula:

\begin{aligned} n & = \log_{1.0781} (\frac{4,211.15}{2,750}) \\ n & = 5.66666... \end{aligned}

$\begin{aligned} n &= \log_{\text{1.0781}}{\left(\frac{\text{4,211.15}}{\text{2,750}}\right)} \\ n &= \text{5.66666...} \end{aligned}$

STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

We know that Wethu left the money in the account for about $\text{5.67}$ years. However, we need to give the answer in terms of years and months.

$\text{5.66666...}$ years $\; = \;$ $\text{5}$ years and $\text{0.66666...}$ of a year (which will be in months).

We know that there are $\text{12}$ months in a year. To convert $\text{0.66666...}$ of a year into months, we do the following calculation:

\begin{matrix} 0.66666... of a year \times (\frac{12 months}{year}) = 8 month(s) \end{matrix}

$\begin{array}{c}\text{0.66666...}\text{ of a year} \times \left( \frac{12\text{ months}}{\text{year}} \right) = \text{8} \text{ month(s)}\end{array}$

Therefore, Wethu deposited the money into the account $\text{5}$ years and $\text{8}$ month(s) ago.

Submit your answer as: and

ID is: 2815 Seed is: 5598

Finding the length of an investment

At some time in the past, Mike deposited $\text{N}\llap{=}\,\text{2,650}$ into a savings account at The Bank of Money Money. The account earns interest at a rate of $\text{7.02}\%$ compounded yearly. How long ago did Mike open the account if the balance is now $\text{N}\llap{=}\,\text{3,193.56}$ ? Give your answer in years and months.

Answer:

The account was opened years and month(s) ago.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

What kind of interest was earned? Which formula do we need for this calculation?

STEP: <no title>
[−0 points ⇒ 4 / 4 points left]

From the question statement, we know that this is a compound interest calculation. We read the question carefully and write down the given information:

\begin{aligned} A & = P (1 + i)^{n} \\ Where: \\ A & = 3,193.56 \\ P & = 2,650 \\ i & = 7.02 % = 0.0702 \\ n & = ? \end{aligned}

$\begin{aligned} A & = P(1+i)^n \\ \text{Where: } & \\ A & = \text{3,193.56} \\ P & = \text{2,650} \\ i & = \text{7.02}\% = \text{0.0702} \\ n & = \text{?} \end{aligned}$

STEP: <no title>
[−2 points ⇒ 2 / 4 points left]

Substitute the known values into the equation and solve for $n$ .

\begin{aligned} 3,193.56 & = 2,650 {(1 + 0.0702)}^{n} \\ 3,193.56 & = 2,650 (1.0702)^{n} \\ \frac{3,193.56}{2,650} & = (1.0702)^{n} \end{aligned}

$\begin{aligned} \text{3,193.56} & = \text{2,650} \left( 1 + \text{0.0702} \right) ^ {n} \\ \text{3,193.56} & = \text{2,650}(\text{1.0702})^{n} \\ \frac{\text{3,193.56}}{\text{2,650}} & = (\text{1.0702})^{n} \end{aligned}$

STEP: <no title>
[−1 point ⇒ 1 / 4 points left]

At this stage, we convert to logarithmic form so that $n$ is the subject of the formula:

\begin{aligned} n & = \log_{1.0702} (\frac{3,193.56}{2,650}) \\ n & = 2.75 \end{aligned}

$\begin{aligned} n &= \log_{\text{1.0702}}{\left(\frac{\text{3,193.56}}{\text{2,650}}\right)} \\ n &= \text{2.75} \end{aligned}$

STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

We know that Mike left the money in the account for about $\text{2.75}$ years. However, we need to give the answer in terms of years and months.

$\text{2.75}$ years $\; = \;$ $\text{2}$ years and $\text{0.75}$ of a year (which will be in months).

We know that there are $\text{12}$ months in a year. To convert $\text{0.75}$ of a year into months, we do the following calculation:

\begin{matrix} 0.75 of a year \times (\frac{12 months}{year}) = 9 month(s) \end{matrix}

$\begin{array}{c}\text{0.75}\text{ of a year} \times \left( \frac{12\text{ months}}{\text{year}} \right) = \text{9} \text{ month(s)}\end{array}$

Therefore, Mike deposited the money into the account $\text{2}$ years and $\text{9}$ month(s) ago.

Submit your answer as: and

ID is: 1532 Seed is: 1525

Timeline question: adding or withdrawing money

Mahikeng City Bank offers a savings account which receives $\text{8.2}\%$ interest p.a. compounded monthly. Khayalethu decides to open an account, and deposits $\text{N}\llap{=}\,\text{6,500}$ . His account accrues interest for $\text{4}$ years until he takes $\text{N}\llap{=}\,\text{3,000}$ from the account. After that, Khayalethu leaves the money in the account until $\text{9}$ years after his original deposit. How much money will be in the account in the end?

Answer: Total amount =

N =

$\text{N}\llap{=}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Read carefully through the question again; write down the things which you know. What is the interest rate, the principal, and so on. Then choose a formula and get started.

STEP: <no title>
[−3 points ⇒ 3 / 6 points left]

This question describes a period of $\text{9}$ years. This period is broken into two sections: $\text{4}$ years after Khayalethu opens his account, and then the last $\text{5}$ years after he takes $\text{N}\llap{=}\,\text{3,000}$ from his account. We must calculate the value in the account separately for these two sections of time.

For the first $\text{4}$ years we have:

\begin{aligned} P & = N = 6,500 \\ i & = 8.2 % = 0.082 \\ n & = 4 \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{N}\llap{=}\,\text{6,500}\\ i &=\text{8.2}\% = \text{0.082}\\ n & = \text{4}\\ A & = \text{ ?} \end{aligned}$

The question tells us that the interest rate is " $\text{8.2}\%$ interest p.a. compounded monthly" - this means that we should use the compound interest formula. We also must remember to multiply the exponent by $\text{12}$ and divide the interest rate by $\text{12}$ because the interest compounds monthly.

\begin{aligned} A & = P (1 + i)^{n} \\ A & = 6,500 {(1 + \frac{0.082}{12})}^{(4 \times 12)} \\ = 6,500 (1.006 . . .)^{48} \\ = 6,500 (1.386 . . .) \\ = N = 9,013.17 ⟵ The amount after 4 years. \end{aligned}

$\begin{aligned} A &= P(1+i)^n \\ A & = \text{6,500} \left( 1 + \frac{\text{0.082}} {\text{12}} \right) ^ {(\text{4} \times \text{12})}\\ & = \text{6,500}(\text{1.006}...)^{\text{48}}\\ & = \text{6,500} (\text{1.386}...)\\ & = \text{N}\llap{=}\,\text{9,013.17} \quad \quad \longleftarrow \text{The amount after } \text{4} \text{ years.} \end{aligned}$

STEP: <no title>
[−1 point ⇒ 2 / 6 points left]

Now we must work out what happens when Khayalethu takes $\text{N}\llap{=}\,\text{3,000}$ from his account: did Khayalethu put more money in, or take money out? You must add or subtract that money now because the amount of money in the account is changing!

\begin{aligned} P & = 9,013.17 - 3,000 = N = 6,013.17 \\ i & = 8.2 % = 0.082 \\ n & = 5 \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{9,013.17} -\text{3,000} = \text{N}\llap{=}\,\text{6,013.17} \\ i &=\text{8.2}\% = \text{0.082}\\ n & = \text{5}\\ A & = \text{ ?} \end{aligned}$

STEP: <no title>
[−2 points ⇒ 0 / 6 points left]

Now solve the second part of the problem, the last $\text{5}$ years. (As above, we must adjust the exponent and the interest rate.)

\begin{aligned} A & = 6,013.17 {(1 + \frac{0.082}{12})}^{(5 \times 12)} \\ = 6,013.17 (1.006 . . .)^{60} \\ = 6,013.17 (1.504 . . .) \\ = N = 9,048.13 ⟵ The amount after 9 years. \end{aligned}

$\begin{aligned} A & = \text{6,013.17} \left( 1 + \frac{\text{0.082}} {\text{12}} \right) ^ {(\text{5} \times \text{12})}\\ & = \text{6,013.17}(\text{1.006}...)^\text{60}\\ & = \text{6,013.17} (\text{1.504}...)\\ & = \text{N}\llap{=}\,\text{9,048.13} \quad \quad \longleftarrow \text{The amount after } \text{9} \text{ years.} \end{aligned}$

In the end, Khayalethu will have $\text{N}\llap{=}\,\text{9,048.13}$ in his account at the bank.

Submit your answer as:

ID is: 1532 Seed is: 71

Timeline question: adding or withdrawing money

Pietermaritzburg Local Bank offers a savings account which gets $\text{6.9}\%$ interest p.a. compounded monthly. Methuli decides to open an account, and deposits $\text{N}\llap{=}\,\text{6,500}$ . His account accrues interest for $\text{4}$ years until he adds $\text{N}\llap{=}\,\text{5,000}$ more into the account. After that, Methuli leaves the money in the account until $\text{6}$ years after his original deposit. How much money will be in the account in the end?

Answer: Total amount =

N =

$\text{N}\llap{=}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Read carefully through the question again; write down the things which you know. What is the interest rate, the principal, and so on. Then choose a formula and get started.

STEP: <no title>
[−3 points ⇒ 3 / 6 points left]

This question describes a period of $\text{6}$ years. This period is broken into two sections: $\text{4}$ years after Methuli opens his account, and then the last $\text{2}$ years after he adds $\text{N}\llap{=}\,\text{5,000}$ more into his account. We must calculate the value in the account separately for these two sections of time.

For the first $\text{4}$ years we have:

\begin{aligned} P & = N = 6,500 \\ i & = 6.9 % = 0.069 \\ n & = 4 \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{N}\llap{=}\,\text{6,500}\\ i &=\text{6.9}\% = \text{0.069}\\ n & = \text{4}\\ A & = \text{ ?} \end{aligned}$

The question tells us that the interest rate is " $\text{6.9}\%$ interest p.a. compounded monthly" - this means that we should use the compound interest formula. We also must remember to multiply the exponent by $\text{12}$ and divide the interest rate by $\text{12}$ because the interest compounds monthly.

\begin{aligned} A & = P (1 + i)^{n} \\ A & = 6,500 {(1 + \frac{0.069}{12})}^{(4 \times 12)} \\ = 6,500 (1.005 . . .)^{48} \\ = 6,500 (1.316 . . .) \\ = N = 8,559.24 ⟵ The amount after 4 years. \end{aligned}

$\begin{aligned} A &= P(1+i)^n \\ A & = \text{6,500} \left( 1 + \frac{\text{0.069}} {\text{12}} \right) ^ {(\text{4} \times \text{12})}\\ & = \text{6,500}(\text{1.005}...)^{\text{48}}\\ & = \text{6,500} (\text{1.316}...)\\ & = \text{N}\llap{=}\,\text{8,559.24} \quad \quad \longleftarrow \text{The amount after } \text{4} \text{ years.} \end{aligned}$

STEP: <no title>
[−1 point ⇒ 2 / 6 points left]

Now we must work out what happens when Methuli adds $\text{N}\llap{=}\,\text{5,000}$ more into his account: did Methuli put more money in, or take money out? You must add or subtract that money now because the amount of money in the account is changing!

\begin{aligned} P & = 8,559.24 +5,000 = N = 13,559.24 \\ i & = 6.9 % = 0.069 \\ n & = 2 \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{8,559.24} \text{+5,000} = \text{N}\llap{=}\,\text{13,559.24} \\ i &=\text{6.9}\% = \text{0.069}\\ n & = \text{2}\\ A & = \text{ ?} \end{aligned}$

STEP: <no title>
[−2 points ⇒ 0 / 6 points left]

Now solve the second part of the problem, the last $\text{2}$ years. (As above, we must adjust the exponent and the interest rate.)

\begin{aligned} A & = 13,559.24 {(1 + \frac{0.069}{12})}^{(2 \times 12)} \\ = 13,559.24 (1.005 . . .)^{24} \\ = 13,559.24 (1.147 . . .) \\ = N = 15,559.53 ⟵ The amount after 6 years. \end{aligned}

$\begin{aligned} A & = \text{13,559.24} \left( 1 + \frac{\text{0.069}} {\text{12}} \right) ^ {(\text{2} \times \text{12})}\\ & = \text{13,559.24}(\text{1.005}...)^\text{24}\\ & = \text{13,559.24} (\text{1.147}...)\\ & = \text{N}\llap{=}\,\text{15,559.53} \quad \quad \longleftarrow \text{The amount after } \text{6} \text{ years.} \end{aligned}$

In the end, Methuli will have $\text{N}\llap{=}\,\text{15,559.53}$ in his account at the bank.

Submit your answer as:

ID is: 1532 Seed is: 3013

Timeline question: adding or withdrawing money

Mahikeng City Bank offers a savings account which receives $\text{9.2}\%$ interest p.a. compounded every week. Lefu decides to open an account, and deposits $\text{N}\llap{=}\,\text{6,500}$ . His account accrues interest for $\text{6}$ years until he withdraws $\text{N}\llap{=}\,\text{2,500}$ from the account. After that, Lefu leaves the money in the account until $\text{13}$ years after his original deposit. How much money will be in the account in the end?

Answer: Total amount =

N =

$\text{N}\llap{=}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Read carefully through the question again; write down the things which you know. What is the interest rate, the principal, and so on. Then choose a formula and get started.

STEP: <no title>
[−3 points ⇒ 3 / 6 points left]

This question describes a period of $\text{13}$ years. This period is broken into two sections: $\text{6}$ years after Lefu opens his account, and then the last $\text{7}$ years after he withdraws $\text{N}\llap{=}\,\text{2,500}$ from his account. We must calculate the value in the account separately for these two sections of time.

For the first $\text{6}$ years we have:

\begin{aligned} P & = N = 6,500 \\ i & = 9.2 % = 0.092 \\ n & = 6 \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{N}\llap{=}\,\text{6,500}\\ i &=\text{9.2}\% = \text{0.092}\\ n & = \text{6}\\ A & = \text{ ?} \end{aligned}$

The question tells us that the interest rate is " $\text{9.2}\%$ interest p.a. compounded every week" - this means that we should use the compound interest formula. We also must remember to multiply the exponent by $\text{52}$ and divide the interest rate by $\text{52}$ because the interest compounds every week.

\begin{aligned} A & = P (1 + i)^{n} \\ A & = 6,500 {(1 + \frac{0.092}{52})}^{(6 \times 52)} \\ = 6,500 (1.001 . . .)^{312} \\ = 6,500 (1.735 . . .) \\ = N = 11,283.19 ⟵ The amount after 6 years. \end{aligned}

$\begin{aligned} A &= P(1+i)^n \\ A & = \text{6,500} \left( 1 + \frac{\text{0.092}} {\text{52}} \right) ^ {(\text{6} \times \text{52})}\\ & = \text{6,500}(\text{1.001}...)^{\text{312}}\\ & = \text{6,500} (\text{1.735}...)\\ & = \text{N}\llap{=}\,\text{11,283.19} \quad \quad \longleftarrow \text{The amount after } \text{6} \text{ years.} \end{aligned}$

STEP: <no title>
[−1 point ⇒ 2 / 6 points left]

Now we must work out what happens when Lefu withdraws $\text{N}\llap{=}\,\text{2,500}$ from his account: did Lefu put more money in, or take money out? You must add or subtract that money now because the amount of money in the account is changing!

\begin{aligned} P & = 11,283.19 - 2,500 = N = 8,783.19 \\ i & = 9.2 % = 0.092 \\ n & = 7 \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{11,283.19} -\text{2,500} = \text{N}\llap{=}\,\text{8,783.19} \\ i &=\text{9.2}\% = \text{0.092}\\ n & = \text{7}\\ A & = \text{ ?} \end{aligned}$

STEP: <no title>
[−2 points ⇒ 0 / 6 points left]

Now solve the second part of the problem, the last $\text{7}$ years. (As above, we must adjust the exponent and the interest rate.)

\begin{aligned} A & = 8,783.19 {(1 + \frac{0.092}{52})}^{(7 \times 52)} \\ = 8,783.19 (1.001 . . .)^{364} \\ = 8,783.19 (1.903 . . .) \\ = N = 16,714.40 ⟵ The amount after 13 years. \end{aligned}

$\begin{aligned} A & = \text{8,783.19} \left( 1 + \frac{\text{0.092}} {\text{52}} \right) ^ {(\text{7} \times \text{52})}\\ & = \text{8,783.19}(\text{1.001}...)^\text{364}\\ & = \text{8,783.19} (\text{1.903}...)\\ & = \text{N}\llap{=}\,\text{16,714.40} \quad \quad \longleftarrow \text{The amount after } \text{13} \text{ years.} \end{aligned}$

In the end, Lefu will have $\text{N}\llap{=}\,\text{16,714.40}$ in his account at the bank.

Submit your answer as:

ID is: 2770 Seed is: 9403

Finding the length of an investment

Simosethu puts $\text{N}\llap{=}\,\text{700}$ into a savings account at the First Bank of Springbok. Simosethu's account earns interest at a rate of $\text{7.24}\%$ p.a. compounded semi-annually. After how many years will the savings account have a balance of $\text{N}\llap{=}\,\text{998.93}$ ?

INSTRUCTION: Give your answer to one decimal place.

Answer: It will take years.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

It is very important to notice that the interest is payable semi-annually, as this affects $i$ and $n$ in the compound interest formula.

STEP: Organise the information given
[−0 points ⇒ 4 / 4 points left]

Write down the known and unknown values:

\begin{aligned} A & = P (1 + i)^{n} \\ Where: \\ A & = 998.93 \\ P & = 700 \\ i & = 7.24 % = 0.0724 \\ n & = ? \end{aligned}

$\begin{aligned} A & = P(1+i)^n \\ \text{Where: } & \\ A & = \text{998.93} \\ P & = \text{700} \\ i & = \text{7.24}\% = \text{0.0724} \\ n & = \text{?} \end{aligned}$

STEP: Adjust the values of

i

$i$ and

n

$n$ based on the compounding period
[−2 points ⇒ 2 / 4 points left]

Note: interest is compounded semi-annually.

We must adjust the values of $i$ and $n$ according to the compounding period. For this question, we must make the following adjustments:

\begin{matrix} i \to \frac{0.0724}{2} and n \to (n \times 2) \end{matrix}

$\begin{array}{c}i \rightarrow \frac{\text{0.0724}}{\text{2}} \quad \text{ and } \quad n \rightarrow (n \times \text{2}) \;\end{array}$

In this case, $n$ represents the number of years and $(n \times \text{2})$ represents the number of times the bank pays interest into the account.

\begin{aligned} 998.93 & = 700 {(1 + \frac{0.0724}{2})}^{(n \times 2)} \end{aligned}

$\begin{aligned} \text{998.93} & = \text{700} \left( 1 + \frac{\text{0.0724}}{\text{2}} \right) ^ {(n \times \text{2})} \end{aligned}$

Simplify the equation:

\begin{aligned} 998.93 & = 700 (1.0362)^{2 n} \\ \frac{998.93}{700} & = (1.0362)^{2 n} \end{aligned}

$\begin{aligned} \text{998.93} & = \text{700}(\text{1.0362})^{\text{2}n} \\ \frac{\text{998.93}}{\text{700}} & = (\text{1.0362})^{\text{2}n} \end{aligned}$

STEP: Rewrite the equation in logarithm form and evaluate the logarithm
[−1 point ⇒ 1 / 4 points left]

Convert the equation to logarithmic form so that $\text{2}n$ becomes the subject of the formula:

\begin{aligned} 2 n & = \log_{1.0362} (\frac{998.93}{700}) \\ 2 n & = 10 \end{aligned}

$\begin{aligned} \text{2}n & = \log_{\text{1.0362}}{\left(\frac{\text{998.93}}{\text{700}}\right)} \\ \text{2}n & = \text{10} \end{aligned}$

STEP: Complete the solution to find the number of years
[−1 point ⇒ 0 / 4 points left]

This is the number of times interest was compounded. To find the number of years, we must solve the equation for $n$ :

\begin{aligned} 2 n & = 10 \\ n & = \frac{10}{2} \\ n & = 5 \end{aligned}

$\begin{aligned} \text{2}n & = \text{10} \\ n & = \frac{\text{10}}{\text{2}} \\ n & = \text{5} \end{aligned}$

Therefore, the money has been in the Simosethu's account for $\text{5}$ years.

NOTE: If you got an answer close to this (for example,

5.12

$\text{5.12}$ years), it probably means that you rounded off at an intermediate step. Avoid rounding off until the end to get the most accurate answer possible.

The number of years is $\text{5}$ .

Submit your answer as:

ID is: 2770 Seed is: 9308

Finding the length of an investment

Asathandwa puts $\text{N}\llap{=}\,\text{550}$ into a bank account at the First Bank of Springbok. Asathandwa's account accrues interest at a rate of $\text{5.6}\%$ p.a. compounded four times per year. After how many years will the bank account have a balance of $\text{N}\llap{=}\,\text{746.79}$ ?

INSTRUCTION: Give your answer to one decimal place.

Answer: It will take years.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

It is very important to notice that the interest is payable four times per year, as this affects $i$ and $n$ in the compound interest formula.

STEP: Organise the information given
[−0 points ⇒ 4 / 4 points left]

Write down the known and unknown values:

\begin{aligned} A & = P (1 + i)^{n} \\ Where: \\ A & = 746.79 \\ P & = 550 \\ i & = 5.6 % = 0.056 \\ n & = ? \end{aligned}

$\begin{aligned} A & = P(1+i)^n \\ \text{Where: } & \\ A & = \text{746.79} \\ P & = \text{550} \\ i & = \text{5.6}\% = \text{0.056} \\ n & = \text{?} \end{aligned}$

STEP: Adjust the values of

i

$i$ and

n

$n$ based on the compounding period
[−2 points ⇒ 2 / 4 points left]

Note: interest is compounded four times per year.

We must adjust the values of $i$ and $n$ according to the compounding period. For this question, we must make the following adjustments:

\begin{matrix} i \to \frac{0.056}{4} and n \to (n \times 4) \end{matrix}

$\begin{array}{c}i \rightarrow \frac{\text{0.056}}{\text{4}} \quad \text{ and } \quad n \rightarrow (n \times \text{4}) \;\end{array}$

In this case, $n$ represents the number of years and $(n \times \text{4})$ represents the number of times the bank pays interest into the account.

\begin{aligned} 746.79 & = 550 {(1 + \frac{0.056}{4})}^{(n \times 4)} \end{aligned}

$\begin{aligned} \text{746.79} & = \text{550} \left( 1 + \frac{\text{0.056}}{\text{4}} \right) ^ {(n \times \text{4})} \end{aligned}$

Simplify the equation:

\begin{aligned} 746.79 & = 550 (1.014)^{4 n} \\ \frac{746.79}{550} & = (1.014)^{4 n} \end{aligned}

$\begin{aligned} \text{746.79} & = \text{550}(\text{1.014})^{\text{4}n} \\ \frac{\text{746.79}}{\text{550}} & = (\text{1.014})^{\text{4}n} \end{aligned}$

STEP: Rewrite the equation in logarithm form and evaluate the logarithm
[−1 point ⇒ 1 / 4 points left]

Convert the equation to logarithmic form so that $\text{4}n$ becomes the subject of the formula:

\begin{aligned} 4 n & = \log_{1.014} (\frac{746.79}{550}) \\ 4 n & = 22 \end{aligned}

$\begin{aligned} \text{4}n & = \log_{\text{1.014}}{\left(\frac{\text{746.79}}{\text{550}}\right)} \\ \text{4}n & = \text{22} \end{aligned}$

STEP: Complete the solution to find the number of years
[−1 point ⇒ 0 / 4 points left]

This is the number of times interest was compounded. To find the number of years, we must solve the equation for $n$ :

\begin{aligned} 4 n & = 22 \\ n & = \frac{22}{4} \\ n & = 5.5 \end{aligned}

$\begin{aligned} \text{4}n & = \text{22} \\ n & = \frac{\text{22}}{\text{4}} \\ n & = \text{5.5} \end{aligned}$

Therefore, the money has been in the Asathandwa's account for $\text{5.5}$ years.

NOTE: If you got an answer close to this (for example,

5.62

$\text{5.62}$ years), it probably means that you rounded off at an intermediate step. Avoid rounding off until the end to get the most accurate answer possible.

The number of years is $\text{5.5}$ .

Submit your answer as:

ID is: 2770 Seed is: 209

Finding the length of an investment

Sive puts $\text{N}\llap{=}\,\text{400}$ into a savings account at the First Bank of Springbok. Sive's account accrues interest at a rate of $\text{7.07}\%$ p.a. compounded every month. After how many years will the savings account have a balance of $\text{N}\llap{=}\,\text{589.44}$ ?

INSTRUCTION: Give your answer to one decimal place.

Answer: It will take years.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

It is very important to notice that the interest is payable every month, as this affects $i$ and $n$ in the compound interest formula.

STEP: Organise the information given
[−0 points ⇒ 4 / 4 points left]

Write down the known and unknown values:

\begin{aligned} A & = P (1 + i)^{n} \\ Where: \\ A & = 589.44 \\ P & = 400 \\ i & = 7.07 % = 0.0707 \\ n & = ? \end{aligned}

$\begin{aligned} A & = P(1+i)^n \\ \text{Where: } & \\ A & = \text{589.44} \\ P & = \text{400} \\ i & = \text{7.07}\% = \text{0.0707} \\ n & = \text{?} \end{aligned}$

STEP: Adjust the values of

i

$i$ and

n

$n$ based on the compounding period
[−2 points ⇒ 2 / 4 points left]

Note: interest is compounded every month.

We must adjust the values of $i$ and $n$ according to the compounding period. For this question, we must make the following adjustments:

\begin{matrix} i \to \frac{0.0707}{12} and n \to (n \times 12) \end{matrix}

$\begin{array}{c}i \rightarrow \frac{\text{0.0707}}{\text{12}} \quad \text{ and } \quad n \rightarrow (n \times \text{12}) \;\end{array}$

In this case, $n$ represents the number of years and $(n \times \text{12})$ represents the number of times the bank pays interest into the account.

\begin{aligned} 589.44 & = 400 {(1 + \frac{0.0707}{12})}^{(n \times 12)} \end{aligned}

$\begin{aligned} \text{589.44} & = \text{400} \left( 1 + \frac{\text{0.0707}}{\text{12}} \right) ^ {(n \times \text{12})} \end{aligned}$

Simplify the equation:

\begin{aligned} 589.44 & = 400 (1.00589...)^{12 n} \\ \frac{589.44}{400} & = (1.00589...)^{12 n} \end{aligned}

$\begin{aligned} \text{589.44} & = \text{400}(\text{1.00589...})^{\text{12}n} \\ \frac{\text{589.44}}{\text{400}} & = (\text{1.00589...})^{\text{12}n} \end{aligned}$

STEP: Rewrite the equation in logarithm form and evaluate the logarithm
[−1 point ⇒ 1 / 4 points left]

Convert the equation to logarithmic form so that $\text{12}n$ becomes the subject of the formula:

\begin{aligned} 12 n & = \log_{1.00589166667} (\frac{589.44}{400}) \\ 12 n & = 66 \end{aligned}

$\begin{aligned} \text{12}n & = \log_{\text{1.00589166667}}{\left(\frac{\text{589.44}}{\text{400}}\right)} \\ \text{12}n & = \text{66} \end{aligned}$

STEP: Complete the solution to find the number of years
[−1 point ⇒ 0 / 4 points left]

This is the number of times interest was compounded. To find the number of years, we must solve the equation for $n$ :

\begin{aligned} 12 n & = 66 \\ n & = \frac{66}{12} \\ n & = 5.5 \end{aligned}

$\begin{aligned} \text{12}n & = \text{66} \\ n & = \frac{\text{66}}{\text{12}} \\ n & = \text{5.5} \end{aligned}$

Therefore, the money has been in the Sive's account for $\text{5.5}$ years.

NOTE: If you got an answer close to this (for example,

5.62

$\text{5.62}$ years), it probably means that you rounded off at an intermediate step. Avoid rounding off until the end to get the most accurate answer possible.

The number of years is $\text{5.5}$ .

Submit your answer as:

ID is: 1530 Seed is: 4943

Timelines

Kamogelo wants to buy a used car. The cost of the used car is $\text{N}\llap{=}\,\text{64,250}$ . In 2004 Kamogelo opened an account at KZN Provincial Bank with $\text{N}\llap{=}\,\text{22,000}$ . Then in 2009 he withdrew $\text{N}\llap{=}\,\text{3,000}$ from the account. In 2015 Kamogelo made another change: he took $\text{N}\llap{=}\,\text{2,500}$ from the account. If the account receives $\text{6.64}\%$ p.a. compounded quarterly, how much money does he have now, and is it enough to buy the used car?

Answer:

The total amount in the account now is R .
Does he have enough money?

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

In a question of this type you must change the years given in the question into the number of years the money was in the account. For example, from 2004 to 2009 is a period of 3 years.

STEP: Find the amount of money there was in 2009 (starting in 2004)
[−2 points ⇒ 5 / 7 points left]

Determine what has been provided and what is required from 2004 to 2009:

\begin{aligned} P & = N = 22,000 \\ i & = 6.64 % = 0.0664 \\ n & = 2009 - 2004 = 5 years \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{N}\llap{=}\,\text{22,000}\\ i &= \text{6.64}\% = \text{0.0664} \\ n & = 2009 - 2004 = \text{5} \text{ years} \\ A & = \text{ ?} \end{aligned}$

Solve the first part of the problem:

\begin{aligned} A & = 22,000 {(1 + \frac{0.0664}{4})}^{(5 \times 4)} \\ A & = 22,000 {(1.0166)}^{20} \\ = 22,000 (1.38995...) \\ = N = 30,579.11 ⟵ The amount in 2009. \end{aligned}

$\begin{aligned} A & = \text{22,000} \left( 1 + \frac{\text{0.0664}} {\text{4}} \right)^ {(\text{5} \times \text{4})}\\ A & = \text{22,000} \left( \text{1.0166} \right)^{\text{20}}\\ & = \text{22,000} \left( \text{1.38995...} \right)\\ & = \text{N}\llap{=}\,\text{30,579.11} \quad \quad \longleftarrow \text{The amount in 2009.} \end{aligned}$

STEP: Adjust the amount of money to reflect Kamogelo's withdrawal from the account
[−2 points ⇒ 3 / 7 points left]

Determine what has been provided and what is required from 2009 to 2015. You must adjust the amount of money in the account if Kamogelo put more money in or took money out.

\begin{aligned} P & = 30,579.11 - 3000 = N = 27,579.11 \\ i & = 6.64 % = 0.0664 \\ n & = 2015 - 2009 = 6 years \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{30,579.11}-3000 = \text{N}\llap{=}\,\text{27,579.11} \\ i &= \text{6.64}\% = \text{0.0664}\\ n & = 2015 - 2009 = \text{6} \text{ years}\\ A & = \text{ ?} \end{aligned}$

Now solve this part of the problem:

\begin{aligned} A & = 27,579.11 {(1 + \frac{0.0664}{4})}^{(6 \times 4)} \\ = 27,579.11 {(1.0166)}^{24} \\ = 27,579.11 (1.48457...) \\ = N = 40,943.29 ⟵ The amount in 2015 . \end{aligned}

$\begin{aligned} A & = \text{27,579.11} \left( 1 + \frac{\text{0.0664}} {\text{4}} \right) ^ {(\text{6} \times \text{4})}\\ & = \text{27,579.11} \left(\text{1.0166} \right)^\text{24}\\ & = \text{27,579.11} \left( \text{1.48457...} \right) \\ & = \text{N}\llap{=}\,\text{40,943.29} \quad \quad \longleftarrow \text{The amount in } 2015 \text{.} \end{aligned}$

STEP: As above, adjust the amount of money in the account
[−2 points ⇒ 1 / 7 points left]

Now from 2015 to 2019: did Kamogelo put more money in, or take money out? As before, you must add or subtract that money now because the amount of money in the account is changing.

\begin{aligned} P & = 40,943.29 - 2500 = N = 38,443.29 \\ i & = 6.64 % = 0.0664 \\ n & = 2019 - 2015 = 4 years \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{40,943.29}-2500 = \text{N}\llap{=}\,\text{38,443.29} \\ i &= \text{6.64}\% = \text{0.0664}\\ n & = 2019 - 2015 = \text{4} \text{ years}\\ A & = \text{ ?} \end{aligned}$

Now solve the third part of the problem, from 2015 to 2019:

\begin{aligned} A & = 38,443.29 {(1 + \frac{0.0664}{4})}^{(4 \times 4)} \\ = 38,443.29 {(1.0166)}^{16} \\ = 38,443.29 (1.30137...) \\ = N = 50,029.05 ⟵ The amount this year. \end{aligned}

$\begin{aligned} A & = \text{38,443.29} \left( 1 + \frac{\text{0.0664}} {\text{4}} \right) ^ {(\text{4} \times \text{4})}\\ & = \text{38,443.29} \left(\text{1.0166} \right)^\text{16} \\ & = \text{38,443.29} \left(\text{1.30137...}\right) \\ & = \text{N}\llap{=}\,\text{50,029.05} \quad \quad \longleftarrow \text{The amount this year.} \end{aligned}$

STEP: State the final answer (including whether or not Kamogelo has enough money)
[−1 point ⇒ 0 / 7 points left]

We have the first answer so the only thing remaining is to determine if there is enough money in the account for Kamogelo to buy the used car.

The answers are:

This year Kamogelo has $\text{N}\llap{=}\,\text{50,029.05}$ in his account, and so
no, he can not afford the used car.

Submit your answer as: and

ID is: 1530 Seed is: 1439

Timelines

Louis wants to buy a motorcycle. The cost of the motorcycle is $\text{N}\llap{=}\,\text{51,750}$ . In 2002 Louis opened an account at First Rand Bank with $\text{N}\llap{=}\,\text{18,000}$ . Then in 2008 he deposited $\text{N}\llap{=}\,\text{3,000}$ more into the account. In 2011 Louis made another change: he withdrew $\text{N}\llap{=}\,\text{4,000}$ from the account. If the account receives $\text{9.17}\%$ p.a. compounded two times each year, how much money does he have now, and is it enough to buy the motorcycle?

Answer:

The total amount in the account now is R .
Does he have enough money?

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

In a question of this type you must change the years given in the question into the number of years the money was in the account. For example, from 2002 to 2008 is a period of 3 years.

STEP: Find the amount of money there was in 2008 (starting in 2002)
[−2 points ⇒ 5 / 7 points left]

Determine what has been provided and what is required from 2002 to 2008:

\begin{aligned} P & = N = 18,000 \\ i & = 9.17 % = 0.0917 \\ n & = 2008 - 2002 = 6 years \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{N}\llap{=}\,\text{18,000}\\ i &= \text{9.17}\% = \text{0.0917} \\ n & = 2008 - 2002 = \text{6} \text{ years} \\ A & = \text{ ?} \end{aligned}$

Solve the first part of the problem:

\begin{aligned} A & = 18,000 {(1 + \frac{0.0917}{2})}^{(6 \times 2)} \\ A & = 18,000 {(1.04585)}^{12} \\ = 18,000 (1.71250...) \\ = N = 30,825.16 ⟵ The amount in 2008. \end{aligned}

$\begin{aligned} A & = \text{18,000} \left( 1 + \frac{\text{0.0917}} {\text{2}} \right)^ {(\text{6} \times \text{2})}\\ A & = \text{18,000} \left( \text{1.04585} \right)^{\text{12}}\\ & = \text{18,000} \left( \text{1.71250...} \right)\\ & = \text{N}\llap{=}\,\text{30,825.16} \quad \quad \longleftarrow \text{The amount in 2008.} \end{aligned}$

STEP: Adjust the amount of money to reflect Louis's deposit into the account
[−2 points ⇒ 3 / 7 points left]

Determine what has been provided and what is required from 2008 to 2011. You must adjust the amount of money in the account if Louis put more money in or took money out.

\begin{aligned} P & = 30,825.16 + 3000 = N = 33,825.16 \\ i & = 9.17 % = 0.0917 \\ n & = 2011 - 2008 = 3 years \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{30,825.16}+3000 = \text{N}\llap{=}\,\text{33,825.16} \\ i &= \text{9.17}\% = \text{0.0917}\\ n & = 2011 - 2008 = \text{3} \text{ years}\\ A & = \text{ ?} \end{aligned}$

Now solve this part of the problem:

\begin{aligned} A & = 33,825.16 {(1 + \frac{0.0917}{2})}^{(3 \times 2)} \\ = 33,825.16 {(1.04585)}^{6} \\ = 33,825.16 (1.30862...) \\ = N = 44,264.57 ⟵ The amount in 2011 . \end{aligned}

$\begin{aligned} A & = \text{33,825.16} \left( 1 + \frac{\text{0.0917}} {\text{2}} \right) ^ {(\text{3} \times \text{2})}\\ & = \text{33,825.16} \left(\text{1.04585} \right)^\text{6}\\ & = \text{33,825.16} \left( \text{1.30862...} \right) \\ & = \text{N}\llap{=}\,\text{44,264.57} \quad \quad \longleftarrow \text{The amount in } 2011 \text{.} \end{aligned}$

STEP: As above, adjust the amount of money in the account
[−2 points ⇒ 1 / 7 points left]

Now from 2011 to 2019: did Louis put more money in, or take money out? As before, you must add or subtract that money now because the amount of money in the account is changing.

\begin{aligned} P & = 44,264.57 - 4000 = N = 40,264.57 \\ i & = 9.17 % = 0.0917 \\ n & = 2019 - 2011 = 8 years \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{44,264.57}-4000 = \text{N}\llap{=}\,\text{40,264.57} \\ i &= \text{9.17}\% = \text{0.0917}\\ n & = 2019 - 2011 = \text{8} \text{ years}\\ A & = \text{ ?} \end{aligned}$

Now solve the third part of the problem, from 2011 to 2019:

\begin{aligned} A & = 40,264.57 {(1 + \frac{0.0917}{2})}^{(8 \times 2)} \\ = 40,264.57 {(1.04585)}^{16} \\ = 40,264.57 (2.04885...) \\ = N = 82,496.11 ⟵ The amount this year. \end{aligned}

$\begin{aligned} A & = \text{40,264.57} \left( 1 + \frac{\text{0.0917}} {\text{2}} \right) ^ {(\text{8} \times \text{2})}\\ & = \text{40,264.57} \left(\text{1.04585} \right)^\text{16} \\ & = \text{40,264.57} \left(\text{2.04885...}\right) \\ & = \text{N}\llap{=}\,\text{82,496.11} \quad \quad \longleftarrow \text{The amount this year.} \end{aligned}$

STEP: State the final answer (including whether or not Louis has enough money)
[−1 point ⇒ 0 / 7 points left]

We have the first answer so the only thing remaining is to determine if there is enough money in the account for Louis to buy the motorcycle.

The answers are:

This year Louis has $\text{N}\llap{=}\,\text{82,496.11}$ in his account, and so
yes, he can afford the motorcycle.

Submit your answer as: and

ID is: 1530 Seed is: 7385

Timelines

Justine wants to buy a used car. The cost of the used car is $\text{N}\llap{=}\,\text{57,500}$ . In 2004 Justine opened an account at Maseru Money Bank with $\text{N}\llap{=}\,\text{18,000}$ . Then in 2007 she withdrew $\text{N}\llap{=}\,\text{2,500}$ from the account. In 2013 Justine made another change: she deposited $\text{N}\llap{=}\,\text{3,800}$ into the account. If the account earns $\text{6.0}\%$ p.a. compounded two times each year, how much money does she have now, and is it enough to buy the used car?

Answer:

The total amount in the account now is R .
Does she have enough money?

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

In a question of this type you must change the years given in the question into the number of years the money was in the account. For example, from 2004 to 2007 is a period of 3 years.

STEP: Find the amount of money there was in 2007 (starting in 2004)
[−2 points ⇒ 5 / 7 points left]

Determine what has been provided and what is required from 2004 to 2007:

\begin{aligned} P & = N = 18,000 \\ i & = 6.0 % = 0.06 \\ n & = 2007 - 2004 = 3 years \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{N}\llap{=}\,\text{18,000}\\ i &= \text{6.0}\% = \text{0.06} \\ n & = 2007 - 2004 = \text{3} \text{ years} \\ A & = \text{ ?} \end{aligned}$

Solve the first part of the problem:

\begin{aligned} A & = 18,000 {(1 + \frac{0.06}{2})}^{(3 \times 2)} \\ A & = 18,000 {(1.03)}^{6} \\ = 18,000 (1.19405...) \\ = N = 21,492.94 ⟵ The amount in 2007. \end{aligned}

$\begin{aligned} A & = \text{18,000} \left( 1 + \frac{\text{0.06}} {\text{2}} \right)^ {(\text{3} \times \text{2})}\\ A & = \text{18,000} \left( \text{1.03} \right)^{\text{6}}\\ & = \text{18,000} \left( \text{1.19405...} \right)\\ & = \text{N}\llap{=}\,\text{21,492.94} \quad \quad \longleftarrow \text{The amount in 2007.} \end{aligned}$

STEP: Adjust the amount of money to reflect Justine's withdrawal from the account
[−2 points ⇒ 3 / 7 points left]

Determine what has been provided and what is required from 2007 to 2013. You must adjust the amount of money in the account if Justine put more money in or took money out.

\begin{aligned} P & = 21,492.94 - 2500 = N = 18,992.94 \\ i & = 6.0 % = 0.06 \\ n & = 2013 - 2007 = 6 years \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{21,492.94}-2500 = \text{N}\llap{=}\,\text{18,992.94} \\ i &= \text{6.0}\% = \text{0.06}\\ n & = 2013 - 2007 = \text{6} \text{ years}\\ A & = \text{ ?} \end{aligned}$

Now solve this part of the problem:

\begin{aligned} A & = 18,992.94 {(1 + \frac{0.06}{2})}^{(6 \times 2)} \\ = 18,992.94 {(1.03)}^{12} \\ = 18,992.94 (1.42576...) \\ = N = 27,079.39 ⟵ The amount in 2013 . \end{aligned}

$\begin{aligned} A & = \text{18,992.94} \left( 1 + \frac{\text{0.06}} {\text{2}} \right) ^ {(\text{6} \times \text{2})}\\ & = \text{18,992.94} \left(\text{1.03} \right)^\text{12}\\ & = \text{18,992.94} \left( \text{1.42576...} \right) \\ & = \text{N}\llap{=}\,\text{27,079.39} \quad \quad \longleftarrow \text{The amount in } 2013 \text{.} \end{aligned}$

STEP: As above, adjust the amount of money in the account
[−2 points ⇒ 1 / 7 points left]

Now from 2013 to 2019: did Justine put more money in, or take money out? As before, you must add or subtract that money now because the amount of money in the account is changing.

\begin{aligned} P & = 27,079.39 + 3800 = N = 30,879.39 \\ i & = 6.0 % = 0.06 \\ n & = 2019 - 2013 = 6 years \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{27,079.39}+3800 = \text{N}\llap{=}\,\text{30,879.39} \\ i &= \text{6.0}\% = \text{0.06}\\ n & = 2019 - 2013 = \text{6} \text{ years}\\ A & = \text{ ?} \end{aligned}$

Now solve the third part of the problem, from 2013 to 2019:

\begin{aligned} A & = 30,879.39 {(1 + \frac{0.06}{2})}^{(6 \times 2)} \\ = 30,879.39 {(1.03)}^{12} \\ = 30,879.39 (1.42576...) \\ = N = 44,026.63 ⟵ The amount this year. \end{aligned}

$\begin{aligned} A & = \text{30,879.39} \left( 1 + \frac{\text{0.06}} {\text{2}} \right) ^ {(\text{6} \times \text{2})}\\ & = \text{30,879.39} \left(\text{1.03} \right)^\text{12} \\ & = \text{30,879.39} \left(\text{1.42576...}\right) \\ & = \text{N}\llap{=}\,\text{44,026.63} \quad \quad \longleftarrow \text{The amount this year.} \end{aligned}$

STEP: State the final answer (including whether or not Justine has enough money)
[−1 point ⇒ 0 / 7 points left]

We have the first answer so the only thing remaining is to determine if there is enough money in the account for Justine to buy the used car.

The answers are:

This year Justine has $\text{N}\llap{=}\,\text{44,026.63}$ in her account, and so
no, she can not afford the used car.

Submit your answer as: and

ID is: 1526 Seed is: 9040

Timeline questions: following the changes

At the beginning of 2007 a young woman starts a savings account at Cape Town Bank. She puts $\text{N}\llap{=}\,\text{5,700}$ into the account. The interest rate is $\text{8.63}$ % p.a. compounded daily. In 2011 the bank changes the interest rate to $\text{7.66}$ % p.a. compounded daily. Then in 2016, the interest rate changes again to $\text{7.66}$ % p.a. compounded weekly.

How much money will the woman have in her account in 2020, $\text{13}$ years after her original deposit?
How much interest will she earn from the bank during this time?

INSTRUCTION: Assume that every year has 365 days (ignore leap years).

Answer:

She will have $\text{N}\llap{=}$ in the account.
She will earn $\text{N}\llap{=}$ in interest.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

Reread the question and write down all of the important information that you know. This is a timeline question: look for where something changes (like an interest rate or a compounding period) to help you break the solution into smaller pieces. For example, in this question, the interest rate changes after $\text{4}$ years.

STEP: <no title>
[−2 points ⇒ 6 / 8 points left]

Determine what has been provided and what is required for the first $\text{4}$ years (from 2007 until 2011):

\begin{aligned} P & = 5,700 \\ i & = 8.63 % = 0.0863 \\ n & = 4 \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{5,700}\\ i &=\text{8.63} \% = \text{0.0863}\\ n & = \text{4}\\ A & = \text{ ?} \end{aligned}$

This is a compound interest problem:

\begin{aligned} A = P (1 + i)^{n} \end{aligned}

$\begin{aligned} \quad A = P(1+i)^n \end{aligned}$

Solve the first part of the problem:

\begin{aligned} A & = 5,700 {(1 + \frac{0.0863}{365})}^{(4 \times 365)} \\ A & = 5,700 (1.000 . . .)^{1,460} \\ = 5,700 (1.412 . . .) \\ = N = 8,049.62 ⟵ The amount after 4 years. \end{aligned}

$\begin{aligned} A & = \text{5,700} \left( 1 + \frac{\text{0.0863}} {\text{365}} \right) ^ {(\text{4} \times \text{365})}\\ A & = \text{5,700}(\text{1.000}...)^{\text{1,460}}\\ & = \text{5,700} (\text{1.412}...)\\ & = \text{N}\llap{=}\,\text{8,049.62} \quad \quad \longleftarrow \text{The amount after } \text{4} \text{ years.} \end{aligned}$

STEP: <no title>
[−2 points ⇒ 4 / 8 points left]

Determine what has been provided and what is required for the next $\text{5}$ years (from 2011 until 2016):

\begin{aligned} P & = N = 8,049.62 \\ i & = 7.66 % = 0.0766 \\ n & = 5 \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{N}\llap{=}\,\text{8,049.62}\\ i &=\text{7.66} \% = \text{0.0766}\\ n & = \text{5}\\ A & = \text{ ?} \end{aligned}$

Now solve the second part of the problem:

\begin{aligned} A & = 8,049.62 {(1 + \frac{0.0766}{365})}^{(5 \times 365)} \\ = 8,049.62 (1.000 . . .)^{1,825} \\ = 8,049.62 (1.466 . . .) \\ = N = 11,805.73 ⟵ The amount after 5 years. \end{aligned}

$\begin{aligned} A & = \text{8,049.62} \left( 1 + \frac{\text{0.0766}} {\text{365}} \right) ^ {(\text{5} \times \text{365})}\\ & = \text{8,049.62}(\text{1.000}...)^\text{1,825}\\ & = \text{8,049.62} (\text{1.466}...)\\ & = \text{N}\llap{=}\,\text{11,805.73} \quad \quad \longleftarrow \text{The amount after } \text{5} \text{ years.} \end{aligned}$

STEP: <no title>
[−2 points ⇒ 2 / 8 points left]

Finally, determine what has been provided and what is required for the next $\text{4}$ years:

\begin{aligned} P & = N = 11,805.73 \\ i & = 7.66 % = 0.0766 \\ n & = 4 \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{N}\llap{=}\,\text{11,805.73}\\ i &=\text{7.66} \% = \text{0.0766}\\ n & = \text{4}\\ A & = \text{ ?} \end{aligned}$

Now solve the third part of the problem, the last $\text{4}$ years:

\begin{aligned} A & = 11,805.73 {(1 + \frac{0.0766}{52})}^{(4 \times 52)} \\ = 11,805.73 (1.001 . . .)^{208} \\ = 11,805.73 (1.358 . . .) \\ = N = 16,034.77 ⟵ The amount after 13 years. \end{aligned}

$\begin{aligned} A & = \text{11,805.73} \left( 1 + \frac{\text{0.0766}} {\text{52}} \right) ^ {(\text{4} \times \text{52})}\\ & = \text{11,805.73}(\text{1.001}...)^\text{208}\\ & = \text{11,805.73} (\text{1.358}...)\\ & = \text{N}\llap{=}\,\text{16,034.77} \quad \quad \longleftarrow \text{The amount after } \text{13} \text{ years.} \end{aligned}$

STEP: <no title>
[−2 points ⇒ 0 / 8 points left]

To calculate the interest earned, subtract the principal amount from the final amount:

\begin{aligned} Interest earned & = 16,034.77 - 5700 \\ = N = 10,334.77 \end{aligned}

$\begin{aligned} \text{Interest earned } & = \text{16,034.77} - 5700 \\ & = \text{N}\llap{=}\,\text{10,334.77} \end{aligned}$

In 2020:

She will have $\text{N}\llap{=}\,\text{16,034.77}$ in her account.
The amount of interest she earns is $\text{N}\llap{=}\,\text{10,334.77}$ .

Submit your answer as: and

ID is: 1526 Seed is: 3398

Timeline questions: following the changes

At the beginning of 2005 a young woman starts a savings account at Mthatha Bank. She invests $\text{N}\llap{=}\,\text{7,350}$ into the account. The interest rate is $\text{8.11}$ % p.a. compounded daily. In 2008 the bank changes the interest rate to $\text{9.0}$ % p.a. compounded daily. Then in 2012, the interest rate changes again to $\text{9.0}$ % p.a. compounded semi-annually.

How much money will the woman have in her account in 2014, $\text{9}$ years after her original deposit?
How much interest will she earn from the bank during this time?

INSTRUCTION: Assume that every year has 365 days (ignore leap years).

Answer:

She will have $\text{N}\llap{=}$ in the account.
She will earn $\text{N}\llap{=}$ in interest.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

STEP: <no title>
[−2 points ⇒ 6 / 8 points left]

Determine what has been provided and what is required for the first $\text{3}$ years (from 2005 until 2008):

\begin{aligned} P & = 7,350 \\ i & = 8.11 % = 0.0811 \\ n & = 3 \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{7,350}\\ i &=\text{8.11} \% = \text{0.0811}\\ n & = \text{3}\\ A & = \text{ ?} \end{aligned}$

This is a compound interest problem:

\begin{aligned} A = P (1 + i)^{n} \end{aligned}

$\begin{aligned} \quad A = P(1+i)^n \end{aligned}$

Solve the first part of the problem:

\begin{aligned} A & = 7,350 {(1 + \frac{0.0811}{365})}^{(3 \times 365)} \\ A & = 7,350 (1.000 . . .)^{1,095} \\ = 7,350 (1.275 . . .) \\ = N = 9,374.31 ⟵ The amount after 3 years. \end{aligned}

$\begin{aligned} A & = \text{7,350} \left( 1 + \frac{\text{0.0811}} {\text{365}} \right) ^ {(\text{3} \times \text{365})}\\ A & = \text{7,350}(\text{1.000}...)^{\text{1,095}}\\ & = \text{7,350} (\text{1.275}...)\\ & = \text{N}\llap{=}\,\text{9,374.31} \quad \quad \longleftarrow \text{The amount after } \text{3} \text{ years.} \end{aligned}$

STEP: <no title>
[−2 points ⇒ 4 / 8 points left]

Determine what has been provided and what is required for the next $\text{4}$ years (from 2008 until 2012):

\begin{aligned} P & = N = 9,374.31 \\ i & = 9.0 % = 0.09 \\ n & = 4 \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{N}\llap{=}\,\text{9,374.31}\\ i &=\text{9.0} \% = \text{0.09}\\ n & = \text{4}\\ A & = \text{ ?} \end{aligned}$

Now solve the second part of the problem:

\begin{aligned} A & = 9,374.31 {(1 + \frac{0.09}{365})}^{(4 \times 365)} \\ = 9,374.31 (1.000 . . .)^{1,460} \\ = 9,374.31 (1.433 . . .) \\ = N = 13,435.88 ⟵ The amount after 4 years. \end{aligned}

$\begin{aligned} A & = \text{9,374.31} \left( 1 + \frac{\text{0.09}} {\text{365}} \right) ^ {(\text{4} \times \text{365})}\\ & = \text{9,374.31}(\text{1.000}...)^\text{1,460}\\ & = \text{9,374.31} (\text{1.433}...)\\ & = \text{N}\llap{=}\,\text{13,435.88} \quad \quad \longleftarrow \text{The amount after } \text{4} \text{ years.} \end{aligned}$

STEP: <no title>
[−2 points ⇒ 2 / 8 points left]

Finally, determine what has been provided and what is required for the next $\text{2}$ years:

\begin{aligned} P & = N = 13,435.88 \\ i & = 9.0 % = 0.09 \\ n & = 2 \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{N}\llap{=}\,\text{13,435.88}\\ i &=\text{9.0} \% = \text{0.09}\\ n & = \text{2}\\ A & = \text{ ?} \end{aligned}$

Now solve the third part of the problem, the last $\text{2}$ years:

\begin{aligned} A & = 13,435.88 {(1 + \frac{0.09}{2})}^{(2 \times 2)} \\ = 13,435.88 (1.045 . . .)^{4} \\ = 13,435.88 (1.192 . . .) \\ = N = 16,022.54 ⟵ The amount after 9 years. \end{aligned}

$\begin{aligned} A & = \text{13,435.88} \left( 1 + \frac{\text{0.09}} {\text{2}} \right) ^ {(\text{2} \times \text{2})}\\ & = \text{13,435.88}(\text{1.045}...)^\text{4}\\ & = \text{13,435.88} (\text{1.192}...)\\ & = \text{N}\llap{=}\,\text{16,022.54} \quad \quad \longleftarrow \text{The amount after } \text{9} \text{ years.} \end{aligned}$

STEP: <no title>
[−2 points ⇒ 0 / 8 points left]

To calculate the interest earned, subtract the principal amount from the final amount:

\begin{aligned} Interest earned & = 16,022.54 - 7350 \\ = N = 8,672.54 \end{aligned}

$\begin{aligned} \text{Interest earned } & = \text{16,022.54} - 7350 \\ & = \text{N}\llap{=}\,\text{8,672.54} \end{aligned}$

In 2014:

She will have $\text{N}\llap{=}\,\text{16,022.54}$ in her account.
The amount of interest she earns is $\text{N}\llap{=}\,\text{8,672.54}$ .

Submit your answer as: and

ID is: 1526 Seed is: 3863

Timeline questions: following the changes

At the beginning of 2010 a young woman starts a savings account at Kimberly Bank. She deposits $\text{N}\llap{=}\,\text{8,250}$ into the account. The interest rate is $\text{6.95}$ % p.a. compounded two times per year. In 2013 the bank changes the interest rate to $\text{6.95}$ % p.a. compounded monthly. Then in 2018, the interest rate changes again to $\text{7.73}$ % p.a. compounded monthly.

How much money will the woman have in her account in 2021, $\text{11}$ years after her original deposit?
How much interest will she earn from the bank during this time?

INSTRUCTION: Assume that every year has 365 days (ignore leap years).

Answer:

She will have $\text{N}\llap{=}$ in the account.
She will earn $\text{N}\llap{=}$ in interest.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

STEP: <no title>
[−2 points ⇒ 6 / 8 points left]

Determine what has been provided and what is required for the first $\text{3}$ years (from 2010 until 2013):

\begin{aligned} P & = 8,250 \\ i & = 6.95 % = 0.0695 \\ n & = 3 \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{8,250}\\ i &=\text{6.95} \% = \text{0.0695}\\ n & = \text{3}\\ A & = \text{ ?} \end{aligned}$

This is a compound interest problem:

\begin{aligned} A = P (1 + i)^{n} \end{aligned}

$\begin{aligned} \quad A = P(1+i)^n \end{aligned}$

Solve the first part of the problem:

\begin{aligned} A & = 8,250 {(1 + \frac{0.0695}{2})}^{(3 \times 2)} \\ A & = 8,250 (1.034 . . .)^{6} \\ = 8,250 (1.227 . . .) \\ = N = 10,126.67 ⟵ The amount after 3 years. \end{aligned}

$\begin{aligned} A & = \text{8,250} \left( 1 + \frac{\text{0.0695}} {\text{2}} \right) ^ {(\text{3} \times \text{2})}\\ A & = \text{8,250}(\text{1.034}...)^{\text{6}}\\ & = \text{8,250} (\text{1.227}...)\\ & = \text{N}\llap{=}\,\text{10,126.67} \quad \quad \longleftarrow \text{The amount after } \text{3} \text{ years.} \end{aligned}$

STEP: <no title>
[−2 points ⇒ 4 / 8 points left]

Determine what has been provided and what is required for the next $\text{5}$ years (from 2013 until 2018):

\begin{aligned} P & = N = 10,126.67 \\ i & = 6.95 % = 0.0695 \\ n & = 5 \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{N}\llap{=}\,\text{10,126.67}\\ i &=\text{6.95} \% = \text{0.0695}\\ n & = \text{5}\\ A & = \text{ ?} \end{aligned}$

Now solve the second part of the problem:

\begin{aligned} A & = 10,126.67 {(1 + \frac{0.0695}{12})}^{(5 \times 12)} \\ = 10,126.67 (1.005 . . .)^{60} \\ = 10,126.67 (1.414 . . .) \\ = N = 14,320.19 ⟵ The amount after 5 years. \end{aligned}

$\begin{aligned} A & = \text{10,126.67} \left( 1 + \frac{\text{0.0695}} {\text{12}} \right) ^ {(\text{5} \times \text{12})}\\ & = \text{10,126.67}(\text{1.005}...)^\text{60}\\ & = \text{10,126.67} (\text{1.414}...)\\ & = \text{N}\llap{=}\,\text{14,320.19} \quad \quad \longleftarrow \text{The amount after } \text{5} \text{ years.} \end{aligned}$

STEP: <no title>
[−2 points ⇒ 2 / 8 points left]

Finally, determine what has been provided and what is required for the next $\text{3}$ years:

\begin{aligned} P & = N = 14,320.19 \\ i & = 7.73 % = 0.0773 \\ n & = 3 \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{N}\llap{=}\,\text{14,320.19}\\ i &=\text{7.73} \% = \text{0.0773}\\ n & = \text{3}\\ A & = \text{ ?} \end{aligned}$

Now solve the third part of the problem, the last $\text{3}$ years:

\begin{aligned} A & = 14,320.19 {(1 + \frac{0.0773}{12})}^{(3 \times 12)} \\ = 14,320.19 (1.006 . . .)^{36} \\ = 14,320.19 (1.260 . . .) \\ = N = 18,044.24 ⟵ The amount after 11 years. \end{aligned}

$\begin{aligned} A & = \text{14,320.19} \left( 1 + \frac{\text{0.0773}} {\text{12}} \right) ^ {(\text{3} \times \text{12})}\\ & = \text{14,320.19}(\text{1.006}...)^\text{36}\\ & = \text{14,320.19} (\text{1.260}...)\\ & = \text{N}\llap{=}\,\text{18,044.24} \quad \quad \longleftarrow \text{The amount after } \text{11} \text{ years.} \end{aligned}$

STEP: <no title>
[−2 points ⇒ 0 / 8 points left]

To calculate the interest earned, subtract the principal amount from the final amount:

\begin{aligned} Interest earned & = 18,044.24 - 8250 \\ = N = 9,794.24 \end{aligned}

$\begin{aligned} \text{Interest earned } & = \text{18,044.24} - 8250 \\ & = \text{N}\llap{=}\,\text{9,794.24} \end{aligned}$

In 2021:

She will have $\text{N}\llap{=}\,\text{18,044.24}$ in her account.
The amount of interest she earns is $\text{N}\llap{=}\,\text{9,794.24}$ .

Submit your answer as: and

ID is: 314 Seed is: 4272

Birthday investment

When his son was $\text{2}$ years old, Chinonso made a deposit of $\text{N}\llap{=}\,\text{1,160}$ in the bank. The investment grew at a simple interest rate and when Chinonso's son was $\text{20}$ years old, the value of the investment was $\text{R}\,\text{2,308.40}$ .

At what rate was the money invested? Give answer correct to one decimal place.

Answer: The simple interest rate is % per annum.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

This is a long word problem. Start by reading through it again and writing down the information you think you need to calculate the answer. Also write down the formula you think you need to answer the question.

STEP: Write down the information given in the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

$A = \text{}\,\text{2,308.40}$
$P = \text{}\,\text{1,160.00}$
$i = ?$
$n = \text{20} - \text{2} = \text{18}$

STEP: Transform the simple interest formula
[−1 point ⇒ 1 / 3 points left]

The question says that the investment "grew at a simple interest rate," so we must use the simple interest formula. To calculate the interest rate, we need to make $i$ the subject of the formula:

\begin{aligned} A & = P (1 + i n) \\ \frac{A}{P} & = 1 + i n \\ \frac{A}{P} - 1 & = i n \\ \frac{\frac{A}{P} - 1}{n} & = i \end{aligned}

$\begin{aligned} A &= P(1+in)\\ \frac{A}{P} &= 1+in\\ \frac{A}{P}-1 &= in\\ \frac{\frac{A}{P}-1}{n} &= i \end{aligned}$

STEP: Substitute and solve
[−1 point ⇒ 0 / 3 points left]

Substitute the known values into the formula and work out the answer.

\begin{aligned} i & = \frac{(\frac{2,308.40}{1,160}) - 1}{18} \\ = 0.055 \\ = 5.5 % per annum \end{aligned}

$\begin{aligned} i &= \frac{\left(\frac{\text{}\,\text{2,308.40}}{\text{1,160}}\right) - 1}{\text{18}}\\ & = \text{0.055}\\ & = \text{5.5}\% \text{ per annum} \end{aligned}$

The simple interest rate is $\text{5.5}\%$ per annum.

Submit your answer as:

ID is: 314 Seed is: 9802

Birthday investment

When his son was $\text{1}$ years old, Jidda made a deposit of $\text{N}\llap{=}\,\text{1,410}$ in the bank. The investment grew at a simple interest rate and when Jidda's son was $\text{7}$ years old, the value of the investment was $\text{R}\,\text{1,841.46}$ .

At what rate was the money invested? Give answer correct to one decimal place.

Answer: The simple interest rate is % per annum.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

STEP: Write down the information given in the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

$A = \text{}\,\text{1,841.46}$
$P = \text{}\,\text{1,410.00}$
$i = ?$
$n = \text{7} - \text{1} = \text{6}$

STEP: Transform the simple interest formula
[−1 point ⇒ 1 / 3 points left]

The question says that the investment "grew at a simple interest rate," so we must use the simple interest formula. To calculate the interest rate, we need to make $i$ the subject of the formula:

\begin{aligned} A & = P (1 + i n) \\ \frac{A}{P} & = 1 + i n \\ \frac{A}{P} - 1 & = i n \\ \frac{\frac{A}{P} - 1}{n} & = i \end{aligned}

$\begin{aligned} A &= P(1+in)\\ \frac{A}{P} &= 1+in\\ \frac{A}{P}-1 &= in\\ \frac{\frac{A}{P}-1}{n} &= i \end{aligned}$

STEP: Substitute and solve
[−1 point ⇒ 0 / 3 points left]

Substitute the known values into the formula and work out the answer.

\begin{aligned} i & = \frac{(\frac{1,841.46}{1,410}) - 1}{6} \\ = 0.051 \\ = 5.1 % per annum \end{aligned}

$\begin{aligned} i &= \frac{\left(\frac{\text{}\,\text{1,841.46}}{\text{1,410}}\right) - 1}{\text{6}}\\ & = \text{0.051}\\ & = \text{5.1}\% \text{ per annum} \end{aligned}$

The simple interest rate is $\text{5.1}\%$ per annum.

Submit your answer as:

ID is: 314 Seed is: 229

Birthday investment

When his son was $\text{3}$ years old, Bongani made a deposit of $\text{N}\llap{=}\,\text{4,170}$ in the bank. The investment grew at a simple interest rate and when Bongani's son was $\text{24}$ years old, the value of the investment was $\text{R}\,\text{8,898.78}$ .

At what rate was the money invested? Give answer correct to one decimal place.

Answer: The simple interest rate is % per annum.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

STEP: Write down the information given in the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

$A = \text{}\,\text{8,898.78}$
$P = \text{}\,\text{4,170.00}$
$i = ?$
$n = \text{24} - \text{3} = \text{21}$

STEP: Transform the simple interest formula
[−1 point ⇒ 1 / 3 points left]

The question says that the investment "grew at a simple interest rate," so we must use the simple interest formula. To calculate the interest rate, we need to make $i$ the subject of the formula:

\begin{aligned} A & = P (1 + i n) \\ \frac{A}{P} & = 1 + i n \\ \frac{A}{P} - 1 & = i n \\ \frac{\frac{A}{P} - 1}{n} & = i \end{aligned}

$\begin{aligned} A &= P(1+in)\\ \frac{A}{P} &= 1+in\\ \frac{A}{P}-1 &= in\\ \frac{\frac{A}{P}-1}{n} &= i \end{aligned}$

STEP: Substitute and solve
[−1 point ⇒ 0 / 3 points left]

Substitute the known values into the formula and work out the answer.

\begin{aligned} i & = \frac{(\frac{8,898.78}{4,170}) - 1}{21} \\ = 0.054 \\ = 5.4 % per annum \end{aligned}

$\begin{aligned} i &= \frac{\left(\frac{\text{}\,\text{8,898.78}}{\text{4,170}}\right) - 1}{\text{21}}\\ & = \text{0.054}\\ & = \text{5.4}\% \text{ per annum} \end{aligned}$

The simple interest rate is $\text{5.4}\%$ per annum.

Submit your answer as:

ID is: 1548 Seed is: 5309

Timeline question: interest earned

Imagine that a friend of yours, Khuthala, deposits $\text{N}\llap{=}\,\text{4,500}$ into a bank account. After $\text{6}$ years the account has a value of $\text{N}\llap{=}\,\text{8,653.26}$ .

What is the total amount of interest the bank pays into her account?

Answer: The interest is R
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
The money in the account at the end come from two different sources: Khuthala, who deposits her own money at the beginning, and the bank, which adds interest into the account. The amount of interest in the account is all of the money except Khuthala's original deposit.

STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

The interest is the amount of money the bank pays into the account; it is different to the interest rate!

The final amount of money in the account is made of the money Khuthala put into the account together with the interest paid into the account. To find the amount of interest the bank paid, you must subtract out any of the money Khuthala put into the account herself.

$\begin{aligned} \textrm{interest earned} & = \text{8,653.26} - \text{4,500} \\ & = \text{4,153.26} \end{aligned}$

Khuthala's account earns $\text{N}\llap{=}\,\text{4,153.26}$ during these $\text{6}$ years.

Submit your answer as:
Now imagine that Khuthala takes $\text{N}\llap{=}\,\text{3,500}$ out of the account $\text{4}$ years after she opened the account. Then after $\text{6}$ years her account has a final value of $\text{N}\llap{=}\,\text{4,300.90}$ . Determine how much interest the bank paid into her account in this situation.

Answer: Interest earned = R
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

Unlike in the first question, the amount of Khuthala's money in the account changes - she takes money from the account. We must deal with this change now when we calculate the interest.

The total amount of money Khuthala contributes to the account is the original deposit of $\text{N}\llap{=}\,\text{4,500}$ minus the $\text{N}\llap{=}\,\text{3,500}$ she took out of the account after $\text{4}$ years. Therefore, in the end, Khuthala has put $\; \text{4,500} -\text{3,500} = \text{1,000} \;$ into the account.

STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

Now we know that $\text{N}\llap{=}\,\text{1,000}$ of the $\text{N}\llap{=}\,\text{4,300.90}$ in the account came from Khuthala's pocket: the rest of it is interest paid into the account by the bank.

So we must do what we did in the first question: subtract Khuthala's money from the amount in the account:

$\begin{aligned} \textrm{interest earned} & = \text{4,300.9} - \text{1,000} \\ & = \text{3,300.9} \end{aligned}$

There we have it: the bank pays $\text{N}\llap{=}\,\text{3,300.90}$ into the account.

Submit your answer as:

ID is: 1548 Seed is: 3395

Timeline question: interest earned

Imagine that a friend of yours, Chidinma, deposits $\text{N}\llap{=}\,\text{4,500}$ into a bank account. After $\text{8}$ years the account has a value of $\text{N}\llap{=}\,\text{9,003.08}$ .

What is the total amount of interest which the account accrues?

Answer: The interest is R
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
The money in the account at the end come from two different sources: Chidinma, who deposits her own money at the beginning, and the bank, which adds interest into the account. The amount of interest in the account is all of the money except Chidinma's original deposit.

STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

The interest is the amount of money the bank pays into the account; it is different to the interest rate!

The final amount of money in the account is made of the money Chidinma put into the account together with the interest paid into the account. To find the amount of interest the bank paid, you must subtract out any of the money Chidinma put into the account herself.

$\begin{aligned} \textrm{interest earned} & = \text{9,003.08} - \text{4,500} \\ & = \text{4,503.08} \end{aligned}$

Chidinma's account earns $\text{N}\llap{=}\,\text{4,503.08}$ during these $\text{8}$ years.

Submit your answer as:
Now imagine that Chidinma puts $\text{N}\llap{=}\,\text{2,500}$ more into the account $\text{5}$ years after she opened the account. Then after $\text{8}$ years her account has a final value of $\text{N}\llap{=}\,\text{12,245.59}$ . Determine how much interest the bank paid into her account in this situation.

Answer: Interest earned = R
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

Unlike in the first question, the amount of Chidinma's money in the account changes - she puts more money into the account. We must deal with this change now when we calculate the interest.

The total amount of money Chidinma contributes to the account is the original deposit of $\text{N}\llap{=}\,\text{4,500}$ plus the $\text{N}\llap{=}\,\text{2,500}$ she added to the account after $\text{5}$ years. Therefore, in the end, Chidinma has put $\; \text{4,500} \text{+2,500} = \text{7,000} \;$ into the account.

STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

Now we know that $\text{N}\llap{=}\,\text{7,000}$ of the $\text{N}\llap{=}\,\text{12,245.59}$ in the account came from Chidinma's pocket: the rest of it is interest paid into the account by the bank.

So we must do what we did in the first question: subtract Chidinma's money from the amount in the account:

$\begin{aligned} \textrm{interest earned} & = \text{12,245.59} - \text{7,000} \\ & = \text{5,245.59} \end{aligned}$

There we have it: the bank pays $\text{N}\llap{=}\,\text{5,245.59}$ into the account.

Submit your answer as:

ID is: 1548 Seed is: 7840

Timeline question: interest earned

Imagine that a friend of yours, Lisa, deposits $\text{N}\llap{=}\,\text{6,000}$ into a bank account. After $\text{6}$ years the account has a value of $\text{N}\llap{=}\,\text{10,415.26}$ .

What is the total amount of interest which the account accrues?

Answer: The interest is R
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
The money in the account at the end come from two different sources: Lisa, who deposits her own money at the beginning, and the bank, which adds interest into the account. The amount of interest in the account is all of the money except Lisa's original deposit.

STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

The interest is the amount of money the bank pays into the account; it is different to the interest rate!

The final amount of money in the account is made of the money Lisa put into the account together with the interest paid into the account. To find the amount of interest the bank paid, you must subtract out any of the money Lisa put into the account herself.

$\begin{aligned} \textrm{interest earned} & = \text{10,415.26} - \text{6,000} \\ & = \text{4,415.26} \end{aligned}$

Lisa's account earns $\text{N}\llap{=}\,\text{4,415.26}$ during these $\text{6}$ years.

Submit your answer as:
Now imagine that Lisa withdraws $\text{N}\llap{=}\,\text{4,500}$ out of the account $\text{4}$ years after she opened the account. Then after $\text{6}$ years her account has a final value of $\text{N}\llap{=}\,\text{5,007.06}$ . Determine how much interest the bank paid into her account in this situation.

Answer: Interest earned = R
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

Unlike in the first question, the amount of Lisa's money in the account changes - she withdraws money from the account. We must deal with this change now when we calculate the interest.

The total amount of money Lisa contributes to the account is the original deposit of $\text{N}\llap{=}\,\text{6,000}$ minus the $\text{N}\llap{=}\,\text{4,500}$ she took out of the account after $\text{4}$ years. Therefore, in the end, Lisa has put $\; \text{6,000} -\text{4,500} = \text{1,500} \;$ into the account.

STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

Now we know that $\text{N}\llap{=}\,\text{1,500}$ of the $\text{N}\llap{=}\,\text{5,007.06}$ in the account came from Lisa's pocket: the rest of it is interest paid into the account by the bank.

So we must do what we did in the first question: subtract Lisa's money from the amount in the account:

$\begin{aligned} \textrm{interest earned} & = \text{5,007.06} - \text{1,500} \\ & = \text{3,507.06} \end{aligned}$

There we have it: the bank pays $\text{N}\llap{=}\,\text{3,507.06}$ into the account.

Submit your answer as:

ID is: 1531 Seed is: 3129

Timelines: following money as it comes and goes

In 2004 Abiodun opens a savings account at Bisho First Bank which gets an interest rate of $\text{9.65}$ % p.a. The interest is compounded every week. Abiodun's original deposit is $\text{N}\llap{=}\,\text{7,250}$ . Then in 2011 he adds $\text{N}\llap{=}\,\text{2,500}$ more into the account. Later, in 2016, he adds $\text{N}\llap{=}\,\text{975}$ into the account.

If the interest rate stays unchanged the entire time, how much money will be in the account in 2018?

Answer: The total amount in the account will be $\text{N}\llap{=}$ .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

This is a long question. Reread the question and try to break it up into smaller pieces. Also think about what you need to do if money is added to the account or if money is taken out of the account.

STEP: <no title>
[−2 points ⇒ 4 / 6 points left]

Determine what has been provided and what is required from 2004 to 2011:

$\begin{aligned} P & = \text{N}\llap{=}\,\text{7,250}\\ i &=\text{9.65} \% = \text{0.0965}\\ n & = 2011 - 2004 = \text{7} \text{ years}\\ A & = \text{ ?} \end{aligned}$

Solve the first part of the problem:

$\begin{aligned} A & = \text{7,250} \left( 1 + \frac{\text{0.0965}} {\text{52}} \right) ^ {(\text{7} \times \text{52})}\\ A & = \text{7,250}(\text{1.001}...)^{\text{364}}\\ & = \text{7,250} (\text{1.963}...)\\ & = \text{N}\llap{=}\,\text{14,237.44} \quad \quad \longleftarrow \text{The amount after } \text{7} \text{ years.} \end{aligned}$

STEP: <no title>
[−2 points ⇒ 2 / 6 points left]

Determine what has been provided and what is required for the next $\text{5}$ years. Did Abiodun put more money in, or take money out? You must add or subtract that money now because the amount of money in the account is changing!

$\begin{aligned} P & = \text{14,237.44}+2500 = \text{N}\llap{=}\,\text{16,737.44} \\ i &=\text{9.65} \% = \text{0.0965}\\ n & = 2016 - 2011 = \text{5} \text{ years}\\ A & = \text{ ?} \end{aligned}$

Now solve the second part of the problem, from 2011 to 2016:

$\begin{aligned} A & = \text{16,737.44} \left( 1 + \frac{\text{0.0965}} {\text{52}} \right) ^ {(\text{5} \times \text{52})}\\ & = \text{16,737.44}(\text{1.001}...)^\text{260}\\ & = \text{16,737.44} (\text{1.619}...)\\ & = \text{N}\llap{=}\,\text{27,104.53} \quad \quad \longleftarrow \text{The amount after } \text{12} \text{ years.} \end{aligned}$

STEP: <no title>
[−2 points ⇒ 0 / 6 points left]

Determine what has been provided and what is required for the next $\text{2}$ years. Did Abiodun put more money in, or take money out? As before, you must add or subtract that money now because the amount of money in the account is changing!

$\begin{aligned} P & = \text{27,104.53}+975 = \text{N}\llap{=}\,\text{28,079.53} \\ i &=\text{9.65} \% = \text{0.0965}\\ n & = 2018 - 2016 = \text{2} \text{ years}\\ A & = \text{ ?} \end{aligned}$

Now solve the third part of the problem, from $\text{2,016}$ to $\text{2,018}$ :

$\begin{aligned} A & = \text{28,079.53} \left( 1 + \frac{\text{0.0965}} {\text{52}} \right) ^ {(\text{2} \times \text{52})}\\ & = \text{28,079.53}(\text{1.001}...)^\text{104}\\ & = \text{28,079.53} (\text{1.212}...)\\ & = \text{N}\llap{=}\,\text{34,051.09} \quad \quad \longleftarrow \text{The amount after } \text{14} \text{ years.} \end{aligned}$

After $\text{14}$ years, Abiodun will have $\text{N}\llap{=}\,\text{34,051.09}$ in his account.

Submit your answer as:
How much interest will the account earn during the entire $\text{14}$ year period?

Answer: The amount of interest he will get is R .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Be careful here! Remember that when you calculate the value of $A$ in the work above, you are finding the total amount of money in the account at the end of the time: this money includes the principal and the interest paid into the account ! You need to find out the interest paid for each of the three separate periods.

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

To find the amount of interest paid into the account takes a bit of care: each of the three calculations above works out the final amount in the account due to interest earned. Therefore, you must subtract that money which went into the account for each calculation from the money that came out at the end of that period:

$\begin{aligned} \text{Interest from the first period} & = \text{14,237.44} - \text{7,250.00} = \text{6,987.44} \\ \text{Interest from the second period} & = \text{27,104.53} - \text{16,737.44} = \text{10,367.09} \\ \text{Interest from the third period} & = \text{34,051.09} - \text{28,079.53} = \text{5,971.56} \\ \text{Total interest}& = \text{N}\llap{=}\,\text{23,326.09} \end{aligned}$

(The total interest is the sum of the three different interest amounts!)

The final answer is: After $\text{14}$ years, the account earns $\text{N}\llap{=}\,\text{23,326.09}$ of interest.

Submit your answer as:

ID is: 1531 Seed is: 2744

Timelines: following money as it comes and goes

In 2007 Adedapo opens a savings account at Masambe Bank which receives an interest rate of $\text{9.77}$ % p.a. The interest is compounded once each month. Adedapo's original deposit is $\text{N}\llap{=}\,\text{5,700}$ . Then in 2011 he takes $\text{N}\llap{=}\,\text{2,200}$ from the account. Later, in 2018, he withdraws $\text{N}\llap{=}\,\text{1,000}$ from the account.

If the interest rate stays unchanged the entire time, how much money will be in the account in 2023?

Answer: The total amount in the account will be $\text{N}\llap{=}$ .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

This is a long question. Reread the question and try to break it up into smaller pieces. Also think about what you need to do if money is added to the account or if money is taken out of the account.

STEP: <no title>
[−2 points ⇒ 4 / 6 points left]

Determine what has been provided and what is required from 2007 to 2011:

$\begin{aligned} P & = \text{N}\llap{=}\,\text{5,700}\\ i &=\text{9.77} \% = \text{0.0977}\\ n & = 2011 - 2007 = \text{4} \text{ years}\\ A & = \text{ ?} \end{aligned}$

Solve the first part of the problem:

$\begin{aligned} A & = \text{5,700} \left( 1 + \frac{\text{0.0977}} {\text{12}} \right) ^ {(\text{4} \times \text{12})}\\ A & = \text{5,700}(\text{1.008}...)^{\text{48}}\\ & = \text{5,700} (\text{1.475}...)\\ & = \text{N}\llap{=}\,\text{8,412.21} \quad \quad \longleftarrow \text{The amount after } \text{4} \text{ years.} \end{aligned}$

STEP: <no title>
[−2 points ⇒ 2 / 6 points left]

Determine what has been provided and what is required for the next $\text{7}$ years. Did Adedapo put more money in, or take money out? You must add or subtract that money now because the amount of money in the account is changing!

$\begin{aligned} P & = \text{8,412.21}-2200 = \text{N}\llap{=}\,\text{6,212.21} \\ i &=\text{9.77} \% = \text{0.0977}\\ n & = 2018 - 2011 = \text{7} \text{ years}\\ A & = \text{ ?} \end{aligned}$

Now solve the second part of the problem, from 2011 to 2018:

$\begin{aligned} A & = \text{6,212.21} \left( 1 + \frac{\text{0.0977}} {\text{12}} \right) ^ {(\text{7} \times \text{12})}\\ & = \text{6,212.21}(\text{1.008}...)^\text{84}\\ & = \text{6,212.21} (\text{1.976}...)\\ & = \text{N}\llap{=}\,\text{12,276.02} \quad \quad \longleftarrow \text{The amount after } \text{11} \text{ years.} \end{aligned}$

STEP: <no title>
[−2 points ⇒ 0 / 6 points left]

Determine what has been provided and what is required for the next $\text{5}$ years. Did Adedapo put more money in, or take money out? As before, you must add or subtract that money now because the amount of money in the account is changing!

$\begin{aligned} P & = \text{12,276.02}-1000 = \text{N}\llap{=}\,\text{11,276.02} \\ i &=\text{9.77} \% = \text{0.0977}\\ n & = 2023 - 2018 = \text{5} \text{ years}\\ A & = \text{ ?} \end{aligned}$

Now solve the third part of the problem, from $\text{2,018}$ to $\text{2,023}$ :

$\begin{aligned} A & = \text{11,276.02} \left( 1 + \frac{\text{0.0977}} {\text{12}} \right) ^ {(\text{5} \times \text{12})}\\ & = \text{11,276.02}(\text{1.008}...)^\text{60}\\ & = \text{11,276.02} (\text{1.626}...)\\ & = \text{N}\llap{=}\,\text{18,342.13} \quad \quad \longleftarrow \text{The amount after } \text{16} \text{ years.} \end{aligned}$

After $\text{16}$ years, Adedapo will have $\text{N}\llap{=}\,\text{18,342.13}$ in his account.

Submit your answer as:
How much interest will the account earn during the entire $\text{16}$ year period?

Answer: The amount of interest he will get is R .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Be careful here! Remember that when you calculate the value of $A$ in the work above, you are finding the total amount of money in the account at the end of the time: this money includes the principal and the interest paid into the account ! You need to find out the interest paid for each of the three separate periods.

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

To find the amount of interest paid into the account takes a bit of care: each of the three calculations above works out the final amount in the account due to interest earned. Therefore, you must subtract that money which went into the account for each calculation from the money that came out at the end of that period:

$\begin{aligned} \text{Interest from the first period} & = \text{8,412.21} - \text{5,700.00} = \text{2,712.21} \\ \text{Interest from the second period} & = \text{12,276.02} - \text{6,212.21} = \text{6,063.81} \\ \text{Interest from the third period} & = \text{18,342.13} - \text{11,276.02} = \text{7,066.11} \\ \text{Total interest}& = \text{N}\llap{=}\,\text{15,842.13} \end{aligned}$

(The total interest is the sum of the three different interest amounts!)

The final answer is: After $\text{16}$ years, the account earns $\text{N}\llap{=}\,\text{15,842.13}$ of interest.

Submit your answer as:

ID is: 1531 Seed is: 7017

Timelines: following money as it comes and goes

In 2005 Isa opens a savings account at Polokwane Local Bank which receives an interest rate of $\text{10.37}$ % p.a. The interest is compounded every week. Isa's original deposit is $\text{N}\llap{=}\,\text{7,050}$ . Then in 2011 he deposits $\text{N}\llap{=}\,\text{2,350}$ more into the account. Later, in 2016, he withdraws $\text{N}\llap{=}\,\text{1,025}$ from the account.

If the interest rate stays unchanged the entire time, how much money will be in the account in 2019?

Answer: The total amount in the account will be $\text{N}\llap{=}$ .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

This is a long question. Reread the question and try to break it up into smaller pieces. Also think about what you need to do if money is added to the account or if money is taken out of the account.

STEP: <no title>
[−2 points ⇒ 4 / 6 points left]

Determine what has been provided and what is required from 2005 to 2011:

$\begin{aligned} P & = \text{N}\llap{=}\,\text{7,050}\\ i &=\text{10.37} \% = \text{0.1037}\\ n & = 2011 - 2005 = \text{6} \text{ years}\\ A & = \text{ ?} \end{aligned}$

Solve the first part of the problem:

$\begin{aligned} A & = \text{7,050} \left( 1 + \frac{\text{0.1037}} {\text{52}} \right) ^ {(\text{6} \times \text{52})}\\ A & = \text{7,050}(\text{1.002}...)^{\text{312}}\\ & = \text{7,050} (\text{1.861}...)\\ & = \text{N}\llap{=}\,\text{13,126.17} \quad \quad \longleftarrow \text{The amount after } \text{6} \text{ years.} \end{aligned}$

STEP: <no title>
[−2 points ⇒ 2 / 6 points left]

Determine what has been provided and what is required for the next $\text{5}$ years. Did Isa put more money in, or take money out? You must add or subtract that money now because the amount of money in the account is changing!

$\begin{aligned} P & = \text{13,126.17}+2350 = \text{N}\llap{=}\,\text{15,476.17} \\ i &=\text{10.37} \% = \text{0.1037}\\ n & = 2016 - 2011 = \text{5} \text{ years}\\ A & = \text{ ?} \end{aligned}$

Now solve the second part of the problem, from 2011 to 2016:

$\begin{aligned} A & = \text{15,476.17} \left( 1 + \frac{\text{0.1037}} {\text{52}} \right) ^ {(\text{5} \times \text{52})}\\ & = \text{15,476.17}(\text{1.002}...)^\text{260}\\ & = \text{15,476.17} (\text{1.678}...)\\ & = \text{N}\llap{=}\,\text{25,978.91} \quad \quad \longleftarrow \text{The amount after } \text{11} \text{ years.} \end{aligned}$

STEP: <no title>
[−2 points ⇒ 0 / 6 points left]

Determine what has been provided and what is required for the next $\text{3}$ years. Did Isa put more money in, or take money out? As before, you must add or subtract that money now because the amount of money in the account is changing!

$\begin{aligned} P & = \text{25,978.91}-1025 = \text{N}\llap{=}\,\text{24,953.91} \\ i &=\text{10.37} \% = \text{0.1037}\\ n & = 2019 - 2016 = \text{3} \text{ years}\\ A & = \text{ ?} \end{aligned}$

Now solve the third part of the problem, from $\text{2,016}$ to $\text{2,019}$ :

$\begin{aligned} A & = \text{24,953.91} \left( 1 + \frac{\text{0.1037}} {\text{52}} \right) ^ {(\text{3} \times \text{52})}\\ & = \text{24,953.91}(\text{1.002}...)^\text{156}\\ & = \text{24,953.91} (\text{1.364}...)\\ & = \text{N}\llap{=}\,\text{34,049.68} \quad \quad \longleftarrow \text{The amount after } \text{14} \text{ years.} \end{aligned}$

After $\text{14}$ years, Isa will have $\text{N}\llap{=}\,\text{34,049.68}$ in his account.

Submit your answer as:
How much interest will the account earn during the entire $\text{14}$ year period?

Answer: The amount of interest he will get is R .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Be careful here! Remember that when you calculate the value of $A$ in the work above, you are finding the total amount of money in the account at the end of the time: this money includes the principal and the interest paid into the account ! You need to find out the interest paid for each of the three separate periods.

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

To find the amount of interest paid into the account takes a bit of care: each of the three calculations above works out the final amount in the account due to interest earned. Therefore, you must subtract that money which went into the account for each calculation from the money that came out at the end of that period:

$\begin{aligned} \text{Interest from the first period} & = \text{13,126.17} - \text{7,050.00} = \text{6,076.17} \\ \text{Interest from the second period} & = \text{25,978.91} - \text{15,476.17} = \text{10,502.74} \\ \text{Interest from the third period} & = \text{34,049.68} - \text{24,953.91} = \text{9,095.77} \\ \text{Total interest}& = \text{N}\llap{=}\,\text{25,674.68} \end{aligned}$

(The total interest is the sum of the three different interest amounts!)

The final answer is: After $\text{14}$ years, the account earns $\text{N}\llap{=}\,\text{25,674.68}$ of interest.

Submit your answer as:

ID is: 316 Seed is: 7290

Savings investment

A person invests an amount of $\text{R}\,\text{4,750}$ in a savings account which pays simple interest at a rate of $\text{8}\%$ per annum.

Calculate the balance accumulated by the end of $\text{3}$ years.

Answer: The final balance is: R .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

You need to use the simple interest formula. Write the formula down and then try to figure out how to use it.

STEP: Organise the information in the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

$A = ?$
$P = \text{}\,\text{4,750}$
$i = \frac{8}{100}$
$n = \text{3}$

STEP: Substitute into the simple interest formula and evaluate
[−2 points ⇒ 0 / 3 points left]

Using the simple interest formula, the accumulated amount is:

\begin{aligned} A & = P (1 + i n) \\ = N = 4,750 (1 + (\frac{8}{100}) \times 3) \\ = R 5,890.00 \end{aligned}

$\begin{aligned} A &= P \left( 1+in \right) \\ &= \text{N}\llap{=}\,\text{4,750} \left(\text{1} + \left(\frac{\text{8}}{\text{100}}\right) \times \text{3}\right) \\ &= \text{R}\,\text{5,890.00} \end{aligned}$

The final balance in the account is $\text{R}\,\text{5,890.00}$ .

Submit your answer as:

ID is: 316 Seed is: 8880

Savings investment

A person invests an amount of $\text{R}\,\text{5,670}$ in a savings account which pays simple interest at a rate of $\text{11}\%$ per annum.

Calculate the balance accumulated by the end of $\text{6}$ years.

Answer: The final balance is: R .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

You need to use the simple interest formula. Write the formula down and then try to figure out how to use it.

STEP: Organise the information in the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

$A = ?$
$P = \text{}\,\text{5,670}$
$i = \frac{11}{100}$
$n = \text{6}$

STEP: Substitute into the simple interest formula and evaluate
[−2 points ⇒ 0 / 3 points left]

Using the simple interest formula, the accumulated amount is:

\begin{aligned} A & = P (1 + i n) \\ = N = 5,670 (1 + (\frac{11}{100}) \times 6) \\ = R 9,412.20 \end{aligned}

$\begin{aligned} A &= P \left( 1+in \right) \\ &= \text{N}\llap{=}\,\text{5,670} \left(\text{1} + \left(\frac{\text{11}}{\text{100}}\right) \times \text{6}\right) \\ &= \text{R}\,\text{9,412.20} \end{aligned}$

The final balance in the account is $\text{R}\,\text{9,412.20}$ .

Submit your answer as:

ID is: 316 Seed is: 8439

Savings investment

A person invests an amount of $\text{R}\,\text{2,800}$ in a savings account which pays simple interest at a rate of $\text{11}\%$ per annum.

Calculate the balance accumulated by the end of $\text{6}$ years.

Answer: The final balance is: R .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

You need to use the simple interest formula. Write the formula down and then try to figure out how to use it.

STEP: Organise the information in the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

$A = ?$
$P = \text{}\,\text{2,800}$
$i = \frac{11}{100}$
$n = \text{6}$

STEP: Substitute into the simple interest formula and evaluate
[−2 points ⇒ 0 / 3 points left]

Using the simple interest formula, the accumulated amount is:

\begin{aligned} A & = P (1 + i n) \\ = N = 2,800 (1 + (\frac{11}{100}) \times 6) \\ = R 4,648.00 \end{aligned}

$\begin{aligned} A &= P \left( 1+in \right) \\ &= \text{N}\llap{=}\,\text{2,800} \left(\text{1} + \left(\frac{\text{11}}{\text{100}}\right) \times \text{6}\right) \\ &= \text{R}\,\text{4,648.00} \end{aligned}$

The final balance in the account is $\text{R}\,\text{4,648.00}$ .

Submit your answer as:

ID is: 1523 Seed is: 8252

Saving for a big purchase

Chris wants to buy a large refrigerator, but right now he doesn't have enough money. A friend told Chris that in $\text{8}\ \text{years}$ the large refrigerator will cost $\text{N}\llap{=}\,\text{8,750}$ . He decides to start saving money today at Limpopo Trust Bank. Chris saves $\text{N}\llap{=}\,\text{5,500}$ into a savings account with an interest rate of $\text{7.86}\%$ p.a. compounded every two months. Then after $\text{2.5}\ \text{years}$ the bank changes the interest rate to $\text{7.3}\%$ p.a. compounded semi-annually. After another $\text{3.5}\ \text{years}$ , the interest rate changes again to $\text{6.92}\%$ p.a. compounded monthly.

How much money will Chris have in the account after $\text{8}\ \text{years}$ , and will he then have enough money to buy the large refrigerator?

Answer:

He will have $\text{R}$ .

Will he have enough money?

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

This question is a beast! When you encounter big questions like this one, try to break the question into smaller pieces. Also, it might be helpful to draw a diagram (a timeline) of what is happening in this situation.

STEP: <no title>
[−2 points ⇒ 5 / 7 points left]

Determine what has been provided and what is required for the first $\text{2.5}\ \text{years}$ :

\begin{aligned} P & = N = 5,500 \\ i & = 7.86 % = 0.0786 \\ n & = 2.5 years \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{N}\llap{=}\,\text{5,500}\\ i &=\text{7.86} \% = \text{0.0786}\\ n & = \text{2.5} \text{ } \text{years}\\ A & = \text{ ?} \end{aligned}$

TIP: Remember to convert

months

$\text{months}$ to

years

$\text{years}$ if necessary!

Solve the first part of the problem:

\begin{aligned} A & = 5,500 {(1 + \frac{0.0786}{6})}^{(2.5 \times 6)} \\ = 5,500 (1.013 . . .)^{15.0} \\ = 5,500 (1.215 . . .) \\ = N = 6,685.71 \end{aligned}

$\begin{aligned} & \\ A & = \text{5,500} \left( 1 + \frac{\text{0.0786}} {\text{6}} \right) ^ {(\text{2.5} \times \text{6})}\\ & = \text{5,500}(\text{1.013}...)^{\text{15.0}}\\ & = \text{5,500} (\text{1.215}...)\\ & = \text{N}\llap{=}\,\text{6,685.71} \end{aligned}$

This is the amount after $\text{2.5}\ \text{years}$ .

STEP: <no title>
[−2 points ⇒ 3 / 7 points left]

Determine what has been provided and what is required for the next $\text{3.5}\ \text{years}$ :

\begin{aligned} P & = N = 6,685.71 \\ i & = 7.3 % = 0.073 \\ n & = 3.5 years \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{N}\llap{=}\,\text{6,685.71}\\ i &=\text{7.3} \% = \text{0.073}\\ n & = \text{3.5} \text{ } \text{years}\\ A & = \text{ ?} \end{aligned}$

TIP: Don't forget to convert

months

$\text{months}$ to

years

$\text{years}$ if necessary!

Now solve the second part of the problem:

\begin{aligned} A & = 6,685.71 {(1 + \frac{0.073}{2})}^{(3.5 \times 2)} \\ = 6,685.71 (1.036 . . .)^{7.0} \\ = 6,685.71 (1.285 . . .) \\ = N = 8,592.76 \end{aligned}

$\begin{aligned} A & = \text{6,685.71} \left( 1 + \frac{\text{0.073}} {\text{2}} \right) ^ {(\text{3.5} \times \text{2})}\\ & = \text{6,685.71}(\text{1.036}...)^\text{7.0}\\ & = \text{6,685.71} (\text{1.285}...)\\ & = \text{N}\llap{=}\,\text{8,592.76} \end{aligned}$

This is the amount after $\text{3.5}\ \text{years}$ .

STEP: <no title>
[−2 points ⇒ 1 / 7 points left]

Determine what has been provided and what is required for the next $\text{2}\ \text{years}$ :

\begin{aligned} P & = N = 8,592.76 \\ i & = 6.92 % = 0.0692 \\ n & = 2 \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{N}\llap{=}\,\text{8,592.76}\\ i &=\text{6.92} \% = \text{0.0692}\\ n & = \text{2}\\ A & = \text{ ?} \end{aligned}$

Now solve the third part of the problem, the last $\text{2}\ \text{years}$ :

\begin{aligned} A & = 8,592.76 {(1 + \frac{0.0692}{12})}^{(2 \times 12)} \\ = 8,592.76 (1.005 . . .)^{24} \\ = 8,592.76 (1.148 . . .) \\ = N = 9,864.30 \end{aligned}

$\begin{aligned} A & = \text{8,592.76} \left( 1 + \frac{\text{0.0692}} {\text{12}} \right) ^ {(\text{2} \times \text{12})}\\ & = \text{8,592.76}(\text{1.005}...)^\text{24}\\ & = \text{8,592.76} (\text{1.148}...)\\ & = \text{N}\llap{=}\,\text{9,864.30} \end{aligned}$

This is the amount after $\text{8}\ \text{years}$ .

STEP: <no title>
[−1 point ⇒ 0 / 7 points left]

Write the final answer, and don't forget to answer both parts of the question!

After $\text{8}$ years, Chris will have $\text{N}\llap{=}\,\text{9,864.30}$ in the account. Yes, he will be able to buy the large refrigerator.

Submit your answer as: and

ID is: 1523 Seed is: 2999

Saving for a big purchase

Fuad wants to buy a huge television, but right now he doesn't have enough money. A friend told Fuad that in $\text{6.5}\ \text{years}$ the huge television will cost $\text{N}\llap{=}\,\text{8,300}$ . He decides to start saving money today at Egoli Bank. Fuad invests $\text{N}\llap{=}\,\text{5,075}$ into a savings account with an interest rate of $\text{5.76}\%$ p.a. compounded two times per year. Then after $\text{18}\ \text{months}$ the bank changes the interest rate to $\text{5.82}\%$ p.a. compounded quarterly. After another $\text{12}\ \text{months}$ , the interest rate changes again to $\text{5.7}\%$ p.a. compounded every two months.

How much money will Fuad have in the account after $\text{6.5}\ \text{years}$ , and will he then have enough money to buy the huge television?

Answer:

He will have $\text{R}$ .

Will he have enough money?

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

STEP: <no title>
[−2 points ⇒ 5 / 7 points left]

Determine what has been provided and what is required for the first $\text{18}\ \text{months}$ :

\begin{aligned} P & = N = 5,075 \\ i & = 5.76 % = 0.0576 \\ n & = 18 months \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{N}\llap{=}\,\text{5,075}\\ i &=\text{5.76} \% = \text{0.0576}\\ n & = \text{18} \text{ } \text{months}\\ A & = \text{ ?} \end{aligned}$

TIP: Remember to convert

months

$\text{months}$ to

years

$\text{years}$ if necessary!

Solve the first part of the problem:

\begin{aligned} A & = 5,075 {(1 + \frac{0.0576}{2})}^{(1.5 \times 2)} \\ = 5,075 (1.028 . . .)^{3.0} \\ = 5,075 (1.088 . . .) \\ = N = 5,526.23 \end{aligned}

$\begin{aligned} & \\ A & = \text{5,075} \left( 1 + \frac{\text{0.0576}} {\text{2}} \right) ^ {(\text{1.5} \times \text{2})}\\ & = \text{5,075}(\text{1.028}...)^{\text{3.0}}\\ & = \text{5,075} (\text{1.088}...)\\ & = \text{N}\llap{=}\,\text{5,526.23} \end{aligned}$

This is the amount after $\text{18}\ \text{months}$ .

STEP: <no title>
[−2 points ⇒ 3 / 7 points left]

Determine what has been provided and what is required for the next $\text{12}\ \text{months}$ :

\begin{aligned} P & = N = 5,526.23 \\ i & = 5.82 % = 0.0582 \\ n & = 12 months \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{N}\llap{=}\,\text{5,526.23}\\ i &=\text{5.82} \% = \text{0.0582}\\ n & = \text{12} \text{ } \text{months}\\ A & = \text{ ?} \end{aligned}$

TIP: Don't forget to convert

months

$\text{months}$ to

years

$\text{years}$ if necessary!

Now solve the second part of the problem:

\begin{aligned} A & = 5,526.23 {(1 + \frac{0.0582}{4})}^{(1.0 \times 4)} \\ = 5,526.23 (1.014 . . .)^{4.0} \\ = 5,526.23 (1.059 . . .) \\ = N = 5,854.94 \end{aligned}

$\begin{aligned} A & = \text{5,526.23} \left( 1 + \frac{\text{0.0582}} {\text{4}} \right) ^ {(\text{1.0} \times \text{4})}\\ & = \text{5,526.23}(\text{1.014}...)^\text{4.0}\\ & = \text{5,526.23} (\text{1.059}...)\\ & = \text{N}\llap{=}\,\text{5,854.94} \end{aligned}$

This is the amount after $\text{12}\ \text{months}$ .

STEP: <no title>
[−2 points ⇒ 1 / 7 points left]

Determine what has been provided and what is required for the next $\text{4}\ \text{years}$ :

\begin{aligned} P & = N = 5,854.94 \\ i & = 5.7 % = 0.057 \\ n & = 4 \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{N}\llap{=}\,\text{5,854.94}\\ i &=\text{5.7} \% = \text{0.057}\\ n & = \text{4}\\ A & = \text{ ?} \end{aligned}$

Now solve the third part of the problem, the last $\text{4}\ \text{years}$ :

\begin{aligned} A & = 5,854.94 {(1 + \frac{0.057}{6})}^{(4 \times 6)} \\ = 5,854.94 (1.009 . . .)^{24} \\ = 5,854.94 (1.254 . . .) \\ = N = 7,346.39 \end{aligned}

$\begin{aligned} A & = \text{5,854.94} \left( 1 + \frac{\text{0.057}} {\text{6}} \right) ^ {(\text{4} \times \text{6})}\\ & = \text{5,854.94}(\text{1.009}...)^\text{24}\\ & = \text{5,854.94} (\text{1.254}...)\\ & = \text{N}\llap{=}\,\text{7,346.39} \end{aligned}$

This is the amount after $\text{6.5}\ \text{years}$ .

STEP: <no title>
[−1 point ⇒ 0 / 7 points left]

Write the final answer, and don't forget to answer both parts of the question!

After $\text{6.5}$ years, Fuad will have $\text{N}\llap{=}\,\text{7,346.39}$ in the account. No, he will not be able to buy the huge television.

Submit your answer as: and

ID is: 1523 Seed is: 8928

Saving for a big purchase

Ndidi wants to buy a large refrigerator, but right now she doesn't have enough money. A friend told Ndidi that in $\text{5.5}\ \text{years}$ the large refrigerator will cost $\text{N}\llap{=}\,\text{8,550}$ . She decides to start saving money today at Port Elizabeth First Bank. Ndidi saves $\text{N}\llap{=}\,\text{5,525}$ into a savings account with an interest rate of $\text{9.99}\%$ p.a. compounded every 3 months. Then after $\text{24}\ \text{months}$ the bank changes the interest rate to $\text{10.46}\%$ p.a. compounded two times per year. After another $\text{18}\ \text{months}$ , the interest rate changes again to $\text{10.77}\%$ p.a. compounded daily.

How much money will Ndidi have in the account after $\text{5.5}\ \text{years}$ , and will she then have enough money to buy the large refrigerator?

Answer:

She will have $\text{R}$ .

Will she have enough money?

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

STEP: <no title>
[−2 points ⇒ 5 / 7 points left]

Determine what has been provided and what is required for the first $\text{24}\ \text{months}$ :

\begin{aligned} P & = N = 5,525 \\ i & = 9.99 % = 0.0999 \\ n & = 24 months \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{N}\llap{=}\,\text{5,525}\\ i &=\text{9.99} \% = \text{0.0999}\\ n & = \text{24} \text{ } \text{months}\\ A & = \text{ ?} \end{aligned}$

TIP: Remember to convert

months

$\text{months}$ to

years

$\text{years}$ if necessary!

Solve the first part of the problem:

\begin{aligned} A & = 5,525 {(1 + \frac{0.0999}{4})}^{(2.0 \times 4)} \\ = 5,525 (1.025 . . .)^{8.0} \\ = 5,525 (1.218 . . .) \\ = N = 6,730.36 \end{aligned}

$\begin{aligned} & \\ A & = \text{5,525} \left( 1 + \frac{\text{0.0999}} {\text{4}} \right) ^ {(\text{2.0} \times \text{4})}\\ & = \text{5,525}(\text{1.025}...)^{\text{8.0}}\\ & = \text{5,525} (\text{1.218}...)\\ & = \text{N}\llap{=}\,\text{6,730.36} \end{aligned}$

This is the amount after $\text{24}\ \text{months}$ .

STEP: <no title>
[−2 points ⇒ 3 / 7 points left]

Determine what has been provided and what is required for the next $\text{18}\ \text{months}$ :

\begin{aligned} P & = N = 6,730.36 \\ i & = 10.46 % = 0.1046 \\ n & = 18 months \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{N}\llap{=}\,\text{6,730.36}\\ i &=\text{10.46} \% = \text{0.1046}\\ n & = \text{18} \text{ } \text{months}\\ A & = \text{ ?} \end{aligned}$

TIP: Don't forget to convert

months

$\text{months}$ to

years

$\text{years}$ if necessary!

Now solve the second part of the problem:

\begin{aligned} A & = 6,730.36 {(1 + \frac{0.1046}{2})}^{(1.5 \times 2)} \\ = 6,730.36 (1.052 . . .)^{3.0} \\ = 6,730.36 (1.165 . . .) \\ = N = 7,842.54 \end{aligned}

$\begin{aligned} A & = \text{6,730.36} \left( 1 + \frac{\text{0.1046}} {\text{2}} \right) ^ {(\text{1.5} \times \text{2})}\\ & = \text{6,730.36}(\text{1.052}...)^\text{3.0}\\ & = \text{6,730.36} (\text{1.165}...)\\ & = \text{N}\llap{=}\,\text{7,842.54} \end{aligned}$

This is the amount after $\text{18}\ \text{months}$ .

STEP: <no title>
[−2 points ⇒ 1 / 7 points left]

Determine what has been provided and what is required for the next $\text{2}\ \text{years}$ :

\begin{aligned} P & = N = 7,842.54 \\ i & = 10.77 % = 0.1077 \\ n & = 2 \\ A & = ? \end{aligned}

$\begin{aligned} P & = \text{N}\llap{=}\,\text{7,842.54}\\ i &=\text{10.77} \% = \text{0.1077}\\ n & = \text{2}\\ A & = \text{ ?} \end{aligned}$

Now solve the third part of the problem, the last $\text{2}\ \text{years}$ :

\begin{aligned} A & = 7,842.54 {(1 + \frac{0.1077}{365})}^{(2 \times 365)} \\ = 7,842.54 (1.000 . . .)^{730} \\ = 7,842.54 (1.240 . . .) \\ = N = 9,727.25 \end{aligned}

$\begin{aligned} A & = \text{7,842.54} \left( 1 + \frac{\text{0.1077}} {\text{365}} \right) ^ {(\text{2} \times \text{365})}\\ & = \text{7,842.54}(\text{1.000}...)^\text{730}\\ & = \text{7,842.54} (\text{1.240}...)\\ & = \text{N}\llap{=}\,\text{9,727.25} \end{aligned}$

This is the amount after $\text{5.5}\ \text{years}$ .

STEP: <no title>
[−1 point ⇒ 0 / 7 points left]

Write the final answer, and don't forget to answer both parts of the question!

After $\text{5.5}$ years, Ndidi will have $\text{N}\llap{=}\,\text{9,727.25}$ in the account. Yes, she will be able to buy the large refrigerator.

Submit your answer as: and

ID is: 315 Seed is: 2016

Simple interest investment

Mario wants to invest $\text{N}\llap{=}\,\text{5,430}$ at a simple interest rate of $\text{12.4}\%$ p.a.

How many years will it take for the money to grow to $\text{N}\llap{=}\,\text{21,177}$ ?

INSTRUCTION: Round up your answer to the nearest year.

Answer: It will take years.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Write down the simple interest formula and the values given in the question. Then use the formula to find the answer.

STEP: Organise the information given in the question
[−1 point ⇒ 3 / 4 points left]

Read the question carefully and write down the given information:

$A = \text{21,177}$
$P = \text{5,430}$
$i = \frac{\text{12.4}}{100}$
$n = ?$

STEP: Transform the simple interest formula for

n

$n$
[−2 points ⇒ 1 / 4 points left]

To calculate the number of years, we need to make $n$ the subject of the formula:

\begin{aligned} A & = P (1 + i n) \\ \frac{A}{P} & = 1 + i n \\ \frac{A}{P} - 1 & = i n \\ \frac{\frac{A}{P} - 1}{i} & = n \end{aligned}

$\begin{aligned} A &= P(1+in)\\ \frac{A}{P}&= 1+in\\ \frac{A}{P}-1 & = in\\ \frac{\frac{A}{P}-1}{i} & = n \end{aligned}$

STEP: Substitute in the values and evalute
[−1 point ⇒ 0 / 4 points left]

Now we substitute the values from the question into the formula. Then evaluate (following BODMAS!) the expression to get the answer.

\begin{aligned} n & = \frac{(\frac{21,177}{5,430}) - 1}{(\frac{12.4}{100})} \\ = 23.38709... \\ = 24 years \Leftarrow round UP to the nearest integer \end{aligned}

$\begin{aligned} n & = \frac{\left(\frac{\text{21,177}}{\text{5,430}}\right) - \text{1}}{\left( \frac{\text{12.4}}{\text{100}} \right)} \\ & = \text{23.38709...}\\ & = \text{24} \text{ years} \quad \Leftarrow \text{round UP to the nearest integer} \end{aligned}$

Rounding up to the nearest year, it will take $\text{24}$ years to reach the goal of saving $\text{N}\llap{=}\,\text{21,177}$ .

The answer, rounded up to the nearest integer, is $\text{24}$ years.

Submit your answer as:

ID is: 315 Seed is: 3263

Simple interest investment

Lefu wants to invest $\text{N}\llap{=}\,\text{3,210}$ at a simple interest rate of $\text{12.1}\%$ p.a.

How many years will it take for the money to grow to $\text{N}\llap{=}\,\text{11,235}$ ?

INSTRUCTION: Round up your answer to the nearest year.

Answer: It will take years.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Write down the simple interest formula and the values given in the question. Then use the formula to find the answer.

STEP: Organise the information given in the question
[−1 point ⇒ 3 / 4 points left]

Read the question carefully and write down the given information:

$A = \text{11,235}$
$P = \text{3,210}$
$i = \frac{\text{12.1}}{100}$
$n = ?$

STEP: Transform the simple interest formula for

n

$n$
[−2 points ⇒ 1 / 4 points left]

To calculate the number of years, we need to make $n$ the subject of the formula:

\begin{aligned} A & = P (1 + i n) \\ \frac{A}{P} & = 1 + i n \\ \frac{A}{P} - 1 & = i n \\ \frac{\frac{A}{P} - 1}{i} & = n \end{aligned}

$\begin{aligned} A &= P(1+in)\\ \frac{A}{P}&= 1+in\\ \frac{A}{P}-1 & = in\\ \frac{\frac{A}{P}-1}{i} & = n \end{aligned}$

STEP: Substitute in the values and evalute
[−1 point ⇒ 0 / 4 points left]

Now we substitute the values from the question into the formula. Then evaluate (following BODMAS!) the expression to get the answer.

\begin{aligned} n & = \frac{(\frac{11,235}{3,210}) - 1}{(\frac{12.1}{100})} \\ = 20.66115... \\ = 21 years \Leftarrow round UP to the nearest integer \end{aligned}

$\begin{aligned} n & = \frac{\left(\frac{\text{11,235}}{\text{3,210}}\right) - \text{1}}{\left( \frac{\text{12.1}}{\text{100}} \right)} \\ & = \text{20.66115...}\\ & = \text{21} \text{ years} \quad \Leftarrow \text{round UP to the nearest integer} \end{aligned}$

Rounding up to the nearest year, it will take $\text{21}$ years to reach the goal of saving $\text{N}\llap{=}\,\text{11,235}$ .

The answer, rounded up to the nearest integer, is $\text{21}$ years.

Submit your answer as:

ID is: 315 Seed is: 834

Simple interest investment

Musa wants to invest $\text{N}\llap{=}\,\text{2,770}$ at a simple interest rate of $\text{10.5}\%$ p.a.

How many years will it take for the money to grow to $\text{N}\llap{=}\,\text{9,418}$ ?

INSTRUCTION: Round up your answer to the nearest year.

Answer: It will take years.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Write down the simple interest formula and the values given in the question. Then use the formula to find the answer.

STEP: Organise the information given in the question
[−1 point ⇒ 3 / 4 points left]

Read the question carefully and write down the given information:

$A = \text{9,418}$
$P = \text{2,770}$
$i = \frac{\text{10.5}}{100}$
$n = ?$

STEP: Transform the simple interest formula for

n

$n$
[−2 points ⇒ 1 / 4 points left]

To calculate the number of years, we need to make $n$ the subject of the formula:

\begin{aligned} A & = P (1 + i n) \\ \frac{A}{P} & = 1 + i n \\ \frac{A}{P} - 1 & = i n \\ \frac{\frac{A}{P} - 1}{i} & = n \end{aligned}

$\begin{aligned} A &= P(1+in)\\ \frac{A}{P}&= 1+in\\ \frac{A}{P}-1 & = in\\ \frac{\frac{A}{P}-1}{i} & = n \end{aligned}$

STEP: Substitute in the values and evalute
[−1 point ⇒ 0 / 4 points left]

Now we substitute the values from the question into the formula. Then evaluate (following BODMAS!) the expression to get the answer.

\begin{aligned} n & = \frac{(\frac{9,418}{2,770}) - 1}{(\frac{10.5}{100})} \\ = 22.85714... \\ = 23 years \Leftarrow round UP to the nearest integer \end{aligned}

$\begin{aligned} n & = \frac{\left(\frac{\text{9,418}}{\text{2,770}}\right) - \text{1}}{\left( \frac{\text{10.5}}{\text{100}} \right)} \\ & = \text{22.85714...}\\ & = \text{23} \text{ years} \quad \Leftarrow \text{round UP to the nearest integer} \end{aligned}$

Rounding up to the nearest year, it will take $\text{23}$ years to reach the goal of saving $\text{N}\llap{=}\,\text{9,418}$ .

The answer, rounded up to the nearest integer, is $\text{23}$ years.

Submit your answer as:

ID is: 311 Seed is: 537

Compound interest rate

Sandile invests $\text{N}\llap{=}\,\text{5,360}$ into an account which pays out a lump sum at the end of $\text{7}$ years.

If he gets $\text{N}\llap{=}\,\text{10,398.40}$ at the end of the period, what effective interest rate did the bank offer him? Give answer correct to one decimal place.

Answer: The interest rate is % per annum.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The question states that the investment has a compound interest rate. Start by writing down the compound interest formula. Then figure out what information you can substitute into the formula.

STEP: Write down the information from the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

$A = \text{}\,\text{10,398.40}$
$P = \text{}\,\text{5,360.00}$
$i = ?$
$n = \text{7}$

STEP: Transform the compound interest formula for

i

$i$
[−1 point ⇒ 1 / 3 points left]

This question is about a compound interest situation, so we need to use the compound interest formula. To calculate the interest rate, we can make $i$ the subject of the formula:

\begin{aligned} A & = P (1 + i)^{n} \\ \frac{A}{P} & = (1 + i)^{n} \\ {(\frac{A}{P})}^{\frac{1}{n}} & = 1 + i \\ {(\frac{A}{P})}^{\frac{1}{n}} - 1 & = i \end{aligned}

$\begin{aligned} A &= P(1+i)^n\\ \frac{A}{P}&= (1+i)^n\\ \left(\frac{A}{P}\right)^{\frac{1}{n}} & = 1+i\\ \left(\frac{A}{P}\right)^{\frac{1}{n}}-1 & = i \end{aligned}$

STEP: Substitute the values and evaluate the answer
[−1 point ⇒ 0 / 3 points left]

Now substitute in the values from above and work out the answer.

\begin{aligned} i & = {(\frac{10,398.40}{5,360})}^{\frac{1}{7}} - 1 \\ = 0.099 \\ = 9.9 % per annum \end{aligned}

$\begin{aligned} i & = \left(\frac{\text{}\,\text{10,398.40}}{\text{5,360}}\right)^{\frac{1}{\text{7}}}-1\\ & = \text{0.099}\\ & = \text{9.9}\% \text{ per annum} \end{aligned}$

For the situation in this question, the effective interest rate is $\text{9.9}\%$ per annum.

Submit your answer as:

ID is: 311 Seed is: 6037

Compound interest rate

Azubuike invests $\text{N}\llap{=}\,\text{6,980}$ into an account which pays out a lump sum at the end of $\text{6}$ years.

If he gets $\text{N}\llap{=}\,\text{10,051.20}$ at the end of the period, what effective interest rate did the bank offer him? Give answer correct to one decimal place.

Answer: The interest rate is % per annum.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The question states that the investment has a compound interest rate. Start by writing down the compound interest formula. Then figure out what information you can substitute into the formula.

STEP: Write down the information from the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

$A = \text{}\,\text{10,051.20}$
$P = \text{}\,\text{6,980.00}$
$i = ?$
$n = \text{6}$

STEP: Transform the compound interest formula for

i

$i$
[−1 point ⇒ 1 / 3 points left]

This question is about a compound interest situation, so we need to use the compound interest formula. To calculate the interest rate, we can make $i$ the subject of the formula:

\begin{aligned} A & = P (1 + i)^{n} \\ \frac{A}{P} & = (1 + i)^{n} \\ {(\frac{A}{P})}^{\frac{1}{n}} & = 1 + i \\ {(\frac{A}{P})}^{\frac{1}{n}} - 1 & = i \end{aligned}

$\begin{aligned} A &= P(1+i)^n\\ \frac{A}{P}&= (1+i)^n\\ \left(\frac{A}{P}\right)^{\frac{1}{n}} & = 1+i\\ \left(\frac{A}{P}\right)^{\frac{1}{n}}-1 & = i \end{aligned}$

STEP: Substitute the values and evaluate the answer
[−1 point ⇒ 0 / 3 points left]

Now substitute in the values from above and work out the answer.

\begin{aligned} i & = {(\frac{10,051.20}{6,980})}^{\frac{1}{6}} - 1 \\ = 0.063 \\ = 6.3 % per annum \end{aligned}

$\begin{aligned} i & = \left(\frac{\text{}\,\text{10,051.20}}{\text{6,980}}\right)^{\frac{1}{\text{6}}}-1\\ & = \text{0.063}\\ & = \text{6.3}\% \text{ per annum} \end{aligned}$

For the situation in this question, the effective interest rate is $\text{6.3}\%$ per annum.

Submit your answer as:

ID is: 311 Seed is: 6247

Compound interest rate

Mandilakhe invests $\text{N}\llap{=}\,\text{4,590}$ into an account which pays out a lump sum at the end of $\text{12}$ years.

If he gets $\text{N}\llap{=}\,\text{8,629.20}$ at the end of the period, what effective interest rate did the bank offer him? Give answer correct to one decimal place.

Answer: The interest rate is % per annum.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The question states that the investment has a compound interest rate. Start by writing down the compound interest formula. Then figure out what information you can substitute into the formula.

STEP: Write down the information from the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

$A = \text{}\,\text{8,629.20}$
$P = \text{}\,\text{4,590.00}$
$i = ?$
$n = \text{12}$

STEP: Transform the compound interest formula for

i

$i$
[−1 point ⇒ 1 / 3 points left]

This question is about a compound interest situation, so we need to use the compound interest formula. To calculate the interest rate, we can make $i$ the subject of the formula:

\begin{aligned} A & = P (1 + i)^{n} \\ \frac{A}{P} & = (1 + i)^{n} \\ {(\frac{A}{P})}^{\frac{1}{n}} & = 1 + i \\ {(\frac{A}{P})}^{\frac{1}{n}} - 1 & = i \end{aligned}

$\begin{aligned} A &= P(1+i)^n\\ \frac{A}{P}&= (1+i)^n\\ \left(\frac{A}{P}\right)^{\frac{1}{n}} & = 1+i\\ \left(\frac{A}{P}\right)^{\frac{1}{n}}-1 & = i \end{aligned}$

STEP: Substitute the values and evaluate the answer
[−1 point ⇒ 0 / 3 points left]

Now substitute in the values from above and work out the answer.

\begin{aligned} i & = {(\frac{8,629.20}{4,590})}^{\frac{1}{12}} - 1 \\ = 0.054 \\ = 5.4 % per annum \end{aligned}

$\begin{aligned} i & = \left(\frac{\text{}\,\text{8,629.20}}{\text{4,590}}\right)^{\frac{1}{\text{12}}}-1\\ & = \text{0.054}\\ & = \text{5.4}\% \text{ per annum} \end{aligned}$

For the situation in this question, the effective interest rate is $\text{5.4}\%$ per annum.

Submit your answer as:

ID is: 1545 Seed is: 1278

Timeline questions: breaking the question down

Read this story, and then answer the questions which follow it.

Save Here Bank offers a savings account which gets $\text{8.7}\%$ interest p.a. compounded every month. Hein decides to open an account, and deposits $\text{N}\llap{=}\,\text{5,750.00}$ . His account accrues interest for $\text{6}$ years until he puts $\text{N}\llap{=}\,\text{2,500}$ more into the account. After that, Hein leaves the money in the account until $\text{10}$ years after his original deposit.

What change happened to the account, and in which year did it happen?

Answer: What changes?
$\qquad$ The change happened after years.
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
Read the story again, and take your time to find information about anything which changes about the account: look for words like: 'change,' 'increase,' 'decrease,' 'adds more money,' 'takes money out,' and so on.

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

There is a lot of information in story about the investment. The key part for this question is: 'His account accrues interest for $\text{6}$ years until he puts $\text{N}\llap{=}\,\text{2,500}$ more into the account.'

During this investment, Hein added money into the account and it happens 6 years after Hein opens the account

Submit your answer as: and
How many years passed from the change in the account to the final year?

Answer: years
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
The investment lasts a total of 10 years, and the change happens 6 years after Hein opens the account. How many years are missing at the end?

STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

The investment lasts a total of 10 years; and from above we know that there was a change 6 years after the account was opened. Then the number of years from the change to the end must be $10 - 6 = 4$ .

It can be very helpful with timeline problems to organise the information with a diagram, like the one shown here. On this diagram, it is easier to see the 4 years at the end of the investment period.

Submit your answer as:
During this investment, how much money did Hein deposit in total?

Answer: R
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
In the beginning, Hein deposits $\text{N}\llap{=}\,\text{5,750.00}$ . How is this amount changed when he puts more into the account?

STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

Hein opens the account with a deposit of $\text{N}\llap{=}\,\text{5,750.00}$ . Later he adds $\text{N}\llap{=}\,\text{2,500}$ more. Therefore, the total amount of money he put into the account is $\text{5,750} + \text{2,500} = \text{8,250}$ .

Submit your answer as:

ID is: 1545 Seed is: 7786

Timeline questions: breaking the question down

Read this story, and then answer the questions which follow it.

Abidemi goes to Save Here Bank and deposits $\text{N}\llap{=}\,\text{3,000.00}$ into an account which pays $\text{11.8}\%$ p.a. compounded semi-annually. After $\text{5}$ years the interest rate changes to $\text{11.8}\%$ p.a., now compounded weekly. Abidemi leaves the money in the account for a total of $\text{12}$ years.

What change happened to the account, and in which year did it happen?

Answer: What changes?
$\qquad$ The change happened after years.
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
Read the story again, and take your time to find information about anything which changes about the account: look for words like: 'change,' 'increase,' 'decrease,' 'adds more money,' 'takes money out,' and so on.

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

There is a lot of information in story about the investment. The key part for this question is: ' After $\text{5}$ years the interest rate changes... now compounded weekly .'

During this investment, the compounding period changed and it happens 5 years after Abidemi opens the account

Submit your answer as: and
How many years passed from the change in the account to the final year?

Answer: years
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
The investment lasts a total of 12 years, and the change happens 5 years after Abidemi opens the account. How many years are missing at the end?

STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

The investment lasts a total of 12 years; and from above we know that there was a change 5 years after the account was opened. Then the number of years from the change to the end must be $12 - 5 = 7$ .

It can be very helpful with timeline problems to organise the information with a diagram, like the one shown here. On this diagram, it is easier to see the 7 years at the end of the investment period.

Submit your answer as:
How often was interest compounded during the 7th year of the investment?

Answer:

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
In the story, the bank makes a change to the interest rate. When does this happen? In the 7th year which interest information applies?

STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

The compounding period for the account changes after 5 years; the 7th year is after the change occurs. Therefore in the 7th year is interest is paid once each week.

Submit your answer as:

ID is: 1545 Seed is: 2239

Timeline questions: breaking the question down

Read this story, and then answer the questions which follow it.

Matthew goes to Save Here Bank and deposits $\text{N}\llap{=}\,\text{6,500.00}$ into an account which pays $\text{10.5}\%$ p.a. compounded monthly. After $\text{5}$ years the bank changes the interest rate to $\text{11.0}\%$ p.a., still compounded monthly. Matthew leaves the money in the account for a total of $\text{8}$ years.

What change happened to the account, and in which year did it happen?

Answer: What changes?
$\qquad$ The change happened after years.
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
Read the story again, and take your time to find information about anything which changes about the account: look for words like: 'change,' 'increase,' 'decrease,' 'adds more money,' 'takes money out,' and so on.

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

There is a lot of information in story about the investment. The key part for this question is: ' After $\text{5}$ years the bank changes the interest rate .'

During this investment, the interest percentage rate changed and it happens 5 years after Matthew opens the account

Submit your answer as: and
How many years passed from the change in the account to the final year?

Answer: years
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
The investment lasts a total of 8 years, and the change happens 5 years after Matthew opens the account. How many years are missing at the end?

STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

The investment lasts a total of 8 years; and from above we know that there was a change 5 years after the account was opened. Then the number of years from the change to the end must be $8 - 5 = 3$ .

It can be very helpful with timeline problems to organise the information with a diagram, like the one shown here. On this diagram, it is easier to see the 3 years at the end of the investment period.

Submit your answer as:
What was the interest rate in the 7th year of the investment?

Answer: %
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
In the story, the bank makes a change to the interest rate. When does this happen? In the 7th year which interest information applies?

STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

The interest rate for the account changes after 5 years; the 7th year is after the change occurs. Therefore the interest rate in the 7th year is the original rate, $\text{11.0}\%$ .

Submit your answer as:

ID is: 2695 Seed is: 2248

Finding the length of an investment

Some years ago, a woman opened a savings account at Limpopo West Bank. The principal amount was $\text{N}\llap{=}\,\text{2,400}$ but now the account has a value of $\text{N}\llap{=}\,\text{5,577.46}$ . For how many years was the money in the account if the account receives $\text{7.28}\%$ compound interest p.a.?

INSTRUCTION: Give your answer to the nearest year.

Answer: The money was in the account for years.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Read the question carefully and organize the given information. Write down the appropriate formula and substitute in the values.

STEP: Organise the information in the question
[−0 points ⇒ 3 / 3 points left]

This is a compound interest calculation, so we write down the formula and organize the given information:

\begin{aligned} A & = P (1 + i)^{n} \\ Where: \\ A & = 5,577.46 \\ P & = 2,400 \\ i & = 7.28 % = 0.0728 \\ n & = ? \end{aligned}

$\begin{aligned} A & = P(1+i)^n \\ \text{Where: } & \\ A & = \text{5,577.46} \\ P & = \text{2,400} \\ i & = \text{7.28}\% = \text{0.0728} \\ n & = \text{?} \end{aligned}$

STEP: Substitute the values into the formula and begin solving for

n

$n$
[−2 points ⇒ 1 / 3 points left]

We need to determine the value of $n$ . Substitute in the known values and make $n$ the subject of the formula:

\begin{aligned} 5,577.46 & = 2,400 {(1 + 0.0728)}^{n} \\ _{this equation by 2,400}^{divide both side of} \to 5,577.46 & = 2,400 (1.0728)^{n} \\ \frac{5,577.46}{2,400} & = (1.0728)^{n} \end{aligned}

$\begin{aligned} \text{5,577.46} & = \text{2,400} \left( 1 + \text{0.0728} \right) ^ {n} \\ ^{\text{divide both side of}}_{\text{this equation by } \text{2,400}} \quad \rightarrow \quad \text{5,577.46} & = \text{2,400}(\text{1.0728})^{n} \\ \frac{\text{5,577.46}}{\text{2,400}} & = (\text{1.0728})^{n} \end{aligned}$

STEP: Rewrite the equation in logarithmic form and calculate

n

$n$
[−1 point ⇒ 0 / 3 points left]

To make $n$ the subject of the formula, we convert the equation from exponential form to logarithmic form:

\begin{aligned} ∴ n & = \log_{1.0728} (\frac{5,577.46}{2,400}) \\ n & = \frac{\log (2.3239...)}{\log 1.0728} \end{aligned}

$\begin{aligned} \therefore n &= \log_{\text{1.0728}}{\left(\frac{\text{5,577.46}}{\text{2,400}}\right)} \\ n &= \frac{\log{\left(\text{2.3239...}\right)}}{\log{\text{1.0728}}} \end{aligned}$

Use a calculator to evaluate the logarithm:

\begin{matrix} n = 12 \end{matrix}

$\begin{array}{c}n = \text{12}\end{array}$

Therefore, the woman left the money in the account for $\text{12}$ years.

Submit your answer as:

ID is: 2695 Seed is: 5167

Finding the length of an investment

Some years ago, a man opened a savings account at Limpopo West Bank. The principal amount was $\text{N}\llap{=}\,\text{3,500}$ but now the account has a value of $\text{N}\llap{=}\,\text{6,063.29}$ . For how many years was the money in the account if the account pays $\text{7.11}\%$ compound interest p.a.?

INSTRUCTION: Give your answer to the nearest year.

Answer: The money was in the account for years.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Read the question carefully and organize the given information. Write down the appropriate formula and substitute in the values.

STEP: Organise the information in the question
[−0 points ⇒ 3 / 3 points left]

This is a compound interest calculation, so we write down the formula and organize the given information:

\begin{aligned} A & = P (1 + i)^{n} \\ Where: \\ A & = 6,063.29 \\ P & = 3,500 \\ i & = 7.11 % = 0.0711 \\ n & = ? \end{aligned}

$\begin{aligned} A & = P(1+i)^n \\ \text{Where: } & \\ A & = \text{6,063.29} \\ P & = \text{3,500} \\ i & = \text{7.11}\% = \text{0.0711} \\ n & = \text{?} \end{aligned}$

STEP: Substitute the values into the formula and begin solving for

n

$n$
[−2 points ⇒ 1 / 3 points left]

We need to determine the value of $n$ . Substitute in the known values and make $n$ the subject of the formula:

\begin{aligned} 6,063.29 & = 3,500 {(1 + 0.0711)}^{n} \\ _{this equation by 3,500}^{divide both side of} \to 6,063.29 & = 3,500 (1.0711)^{n} \\ \frac{6,063.29}{3,500} & = (1.0711)^{n} \end{aligned}

$\begin{aligned} \text{6,063.29} & = \text{3,500} \left( 1 + \text{0.0711} \right) ^ {n} \\ ^{\text{divide both side of}}_{\text{this equation by } \text{3,500}} \quad \rightarrow \quad \text{6,063.29} & = \text{3,500}(\text{1.0711})^{n} \\ \frac{\text{6,063.29}}{\text{3,500}} & = (\text{1.0711})^{n} \end{aligned}$

STEP: Rewrite the equation in logarithmic form and calculate

n

$n$
[−1 point ⇒ 0 / 3 points left]

To make $n$ the subject of the formula, we convert the equation from exponential form to logarithmic form:

\begin{aligned} ∴ n & = \log_{1.0711} (\frac{6,063.29}{3,500}) \\ n & = \frac{\log (1.7323...)}{\log 1.0711} \end{aligned}

$\begin{aligned} \therefore n &= \log_{\text{1.0711}}{\left(\frac{\text{6,063.29}}{\text{3,500}}\right)} \\ n &= \frac{\log{\left(\text{1.7323...}\right)}}{\log{\text{1.0711}}} \end{aligned}$

Use a calculator to evaluate the logarithm:

\begin{matrix} n = 8 \end{matrix}

$\begin{array}{c}n = \text{8}\end{array}$

Therefore, the man left the money in the account for $\text{8}$ years.

Submit your answer as:

ID is: 2695 Seed is: 7195

Finding the length of an investment

Some years ago, a woman opened a savings account at Limpopo West Bank. The principal amount was $\text{N}\llap{=}\,\text{3,300}$ but now the account has a value of $\text{N}\llap{=}\,\text{7,935.00}$ . For how many years was the money in the account if the account gets $\text{9.17}\%$ compound interest p.a.?

INSTRUCTION: Give your answer to the nearest year.

Answer: The money was in the account for years.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Read the question carefully and organize the given information. Write down the appropriate formula and substitute in the values.

STEP: Organise the information in the question
[−0 points ⇒ 3 / 3 points left]

This is a compound interest calculation, so we write down the formula and organize the given information:

\begin{aligned} A & = P (1 + i)^{n} \\ Where: \\ A & = 7,935.0 \\ P & = 3,300 \\ i & = 9.17 % = 0.0917 \\ n & = ? \end{aligned}

$\begin{aligned} A & = P(1+i)^n \\ \text{Where: } & \\ A & = \text{7,935.0} \\ P & = \text{3,300} \\ i & = \text{9.17}\% = \text{0.0917} \\ n & = \text{?} \end{aligned}$

STEP: Substitute the values into the formula and begin solving for

n

$n$
[−2 points ⇒ 1 / 3 points left]

We need to determine the value of $n$ . Substitute in the known values and make $n$ the subject of the formula:

\begin{aligned} 7,935.0 & = 3,300 {(1 + 0.0917)}^{n} \\ _{this equation by 3,300}^{divide both side of} \to 7,935.0 & = 3,300 (1.0917)^{n} \\ \frac{7,935.0}{3,300} & = (1.0917)^{n} \end{aligned}

$\begin{aligned} \text{7,935.0} & = \text{3,300} \left( 1 + \text{0.0917} \right) ^ {n} \\ ^{\text{divide both side of}}_{\text{this equation by } \text{3,300}} \quad \rightarrow \quad \text{7,935.0} & = \text{3,300}(\text{1.0917})^{n} \\ \frac{\text{7,935.0}}{\text{3,300}} & = (\text{1.0917})^{n} \end{aligned}$

STEP: Rewrite the equation in logarithmic form and calculate

n

$n$
[−1 point ⇒ 0 / 3 points left]

To make $n$ the subject of the formula, we convert the equation from exponential form to logarithmic form:

\begin{aligned} ∴ n & = \log_{1.0917} (\frac{7,935.0}{3,300}) \\ n & = \frac{\log (2.4045...)}{\log 1.0917} \end{aligned}

$\begin{aligned} \therefore n &= \log_{\text{1.0917}}{\left(\frac{\text{7,935.0}}{\text{3,300}}\right)} \\ n &= \frac{\log{\left(\text{2.4045...}\right)}}{\log{\text{1.0917}}} \end{aligned}$

Use a calculator to evaluate the logarithm:

\begin{matrix} n = 10 \end{matrix}

$\begin{array}{c}n = \text{10}\end{array}$

Therefore, the woman left the money in the account for $\text{10}$ years.

Submit your answer as:

ID is: 318 Seed is: 5834

Compound interest investment

Aisha wants to invest some money at a compound interest rate of $\text{7.9}\%$ p.a.

How much money should be invested if she wants to reach a sum of $\text{N}\llap{=}\,\text{67,300}$ in $\text{3}$ years' time?

INSTRUCTION: Round up your answer to the nearest rand.

Answer: Aisha must invest R .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The question states that the investment has a compound interest rate. Start by writing down the compound interest formula. Then figure out what information you can substitute into the formula.

STEP: Write down the information from the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

$A = \text{N}\llap{=}\,\text{67,300}$
$P = ?$
$i = \text{7.9}\% = \frac{\text{7.9}}{100}$
$n = \text{3}$

STEP: Transform the equation for

P

$P$
[−1 point ⇒ 1 / 3 points left]

The question states that the investment grows at a compound interest rate. So we will use the compound interest formula. To determine the amount she must invest, we can to make $P$ the subject of the formula:

\begin{aligned} A & = P (1 + i)^{n} \\ \frac{A}{(1 + i)^{n}} & = P \end{aligned}

$\begin{aligned} A &= P(1+i)^n\\ \frac{A}{(1+i)^n} &= P \end{aligned}$

STEP: Substitute in the values and work out the answer
[−1 point ⇒ 0 / 3 points left]

\begin{aligned} \frac{67,300}{{(1 + \frac{7.9}{100})}^{3}} & = P \\ \frac{67,300}{{(1.079)}^{3}} & = P \\ P = & N = 53,573.59 \end{aligned}

$\begin{aligned} \frac{\text{}\,\text{67,300}}{\left(1+\frac{\text{7.9}}{100} \right)^{\text{3}}} &= P\\ \frac{\text{}\,\text{67,300}}{\left(\text{1.079} \right)^{\text{3}}} &= P\\ P = & \text{N}\llap{=}\,\text{53,573.59} \end{aligned}$

The result of the calculation shows that Aisha must invest $\text{N}\llap{=}\,\text{53,573.59}$ if she is going to reach her goal of $\text{N}\llap{=}\,\text{67,300}$ in $\text{3}$ years. The question tells us to "round up your answer to the nearest rand," so the final answer is $\text{N}\llap{=}\,\text{53,574}$ .

NOTE:

Rounding up is important here. In this question, Aisha is aiming to reach a final goal of $\text{N}\llap{=}\,\text{67,300}$ . If we round the answer down, then after $\text{3}$ years the total amount of money will not quite reach $\text{N}\llap{=}\,\text{67,300}$ - it will be very close, but not equal to the final goal. In a situation like this, it is common to round up instead.

Rounding up to the nearest cent would work, but we were asked to round up to the nearest rand.

The correct answer is $\text{N}\llap{=}\,\text{53,574}$ .

Submit your answer as:

ID is: 318 Seed is: 6855

Compound interest investment

Nkechi wants to invest some money at a compound interest rate of $\text{8.0}\%$ p.a.

How much money should be invested if she wants to reach a sum of $\text{N}\llap{=}\,\text{16,800}$ in $\text{7}$ years' time?

INSTRUCTION: Round up your answer to the nearest rand.

Answer: Nkechi must invest R .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The question states that the investment has a compound interest rate. Start by writing down the compound interest formula. Then figure out what information you can substitute into the formula.

STEP: Write down the information from the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

$A = \text{N}\llap{=}\,\text{16,800}$
$P = ?$
$i = \text{8.0}\% = \frac{\text{8.0}}{100}$
$n = \text{7}$

STEP: Transform the equation for

P

$P$
[−1 point ⇒ 1 / 3 points left]

\begin{aligned} A & = P (1 + i)^{n} \\ \frac{A}{(1 + i)^{n}} & = P \end{aligned}

$\begin{aligned} A &= P(1+i)^n\\ \frac{A}{(1+i)^n} &= P \end{aligned}$

STEP: Substitute in the values and work out the answer
[−1 point ⇒ 0 / 3 points left]

\begin{aligned} \frac{16,800}{{(1 + \frac{8.0}{100})}^{7}} & = P \\ \frac{16,800}{{(1.08)}^{7}} & = P \\ P = & N = 9,802.64 \end{aligned}

$\begin{aligned} \frac{\text{}\,\text{16,800}}{\left(1+\frac{\text{8.0}}{100} \right)^{\text{7}}} &= P\\ \frac{\text{}\,\text{16,800}}{\left(\text{1.08} \right)^{\text{7}}} &= P\\ P = & \text{N}\llap{=}\,\text{9,802.64} \end{aligned}$

The result of the calculation shows that Nkechi must invest $\text{N}\llap{=}\,\text{9,802.64}$ if she is going to reach her goal of $\text{N}\llap{=}\,\text{16,800}$ in $\text{7}$ years. The question tells us to "round up your answer to the nearest rand," so the final answer is $\text{N}\llap{=}\,\text{9,803}$ .

NOTE:

Rounding up is important here. In this question, Nkechi is aiming to reach a final goal of $\text{N}\llap{=}\,\text{16,800}$ . If we round the answer down, then after $\text{7}$ years the total amount of money will not quite reach $\text{N}\llap{=}\,\text{16,800}$ - it will be very close, but not equal to the final goal. In a situation like this, it is common to round up instead.

Rounding up to the nearest cent would work, but we were asked to round up to the nearest rand.

The correct answer is $\text{N}\llap{=}\,\text{9,803}$ .

Submit your answer as:

ID is: 318 Seed is: 7521

Compound interest investment

Justine wants to invest some money at a compound interest rate of $\text{14.0}\%$ p.a.

How much money should be invested if she wants to reach a sum of $\text{N}\llap{=}\,\text{40,500}$ in $\text{5}$ years' time?

INSTRUCTION: Round up your answer to the nearest rand.

Answer: Justine must invest R .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The question states that the investment has a compound interest rate. Start by writing down the compound interest formula. Then figure out what information you can substitute into the formula.

STEP: Write down the information from the question
[−1 point ⇒ 2 / 3 points left]

Read the question carefully and write down the given information:

$A = \text{N}\llap{=}\,\text{40,500}$
$P = ?$
$i = \text{14.0}\% = \frac{\text{14.0}}{100}$
$n = \text{5}$

STEP: Transform the equation for

P

$P$
[−1 point ⇒ 1 / 3 points left]

\begin{aligned} A & = P (1 + i)^{n} \\ \frac{A}{(1 + i)^{n}} & = P \end{aligned}

$\begin{aligned} A &= P(1+i)^n\\ \frac{A}{(1+i)^n} &= P \end{aligned}$

STEP: Substitute in the values and work out the answer
[−1 point ⇒ 0 / 3 points left]

\begin{aligned} \frac{40,500}{{(1 + \frac{14.0}{100})}^{5}} & = P \\ \frac{40,500}{{(1.14)}^{5}} & = P \\ P = & N = 21,034.43 \end{aligned}

$\begin{aligned} \frac{\text{}\,\text{40,500}}{\left(1+\frac{\text{14.0}}{100} \right)^{\text{5}}} &= P\\ \frac{\text{}\,\text{40,500}}{\left(\text{1.14} \right)^{\text{5}}} &= P\\ P = & \text{N}\llap{=}\,\text{21,034.43} \end{aligned}$

The result of the calculation shows that Justine must invest $\text{N}\llap{=}\,\text{21,034.43}$ if she is going to reach her goal of $\text{N}\llap{=}\,\text{40,500}$ in $\text{5}$ years. The question tells us to "round up your answer to the nearest rand," so the final answer is $\text{N}\llap{=}\,\text{21,035}$ .

NOTE:

Rounding up is important here. In this question, Justine is aiming to reach a final goal of $\text{N}\llap{=}\,\text{40,500}$ . If we round the answer down, then after $\text{5}$ years the total amount of money will not quite reach $\text{N}\llap{=}\,\text{40,500}$ - it will be very close, but not equal to the final goal. In a situation like this, it is common to round up instead.

Rounding up to the nearest cent would work, but we were asked to round up to the nearest rand.

The correct answer is $\text{N}\llap{=}\,\text{21,035}$ .

Submit your answer as:

ID is: 1521 Seed is: 148

Timeline questions: changing interest rates

Emma goes to We Pay More Bank and deposits $\text{N}\llap{=}\,\text{3,000.00}$ into an account which pays $\text{8.8}\%$ p.a. compounded every six months. After $\text{6}$ years the bank changes the interest rate to $\text{10.1}\%$ p.a., still compounded every six months.

How much money will Emma have in her account $\text{10}$ years after the original investment?

Answer: Total amount = $\text{N}\llap{=}$
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

This question is a timeline question in which the interest rate changes. When does it change? How many times does it change? You need to break the question into separate pieces between each of the interest rate changes, and work through each of those pieces to get the final answer.

STEP: <no title>
[−3 points ⇒ 3 / 6 points left]

This question describes a $\text{10}$ year period which is broken into two sections because of the change in the interest rate: the first section lasts $\text{6}$ years (while the interest rate is $\text{8.8}\%$ ); the second section lasts $\text{4}$ years (when the rate is $\text{10.1}\%$ ).

Determine what has been provided and what is required for the first $\text{6}$ years:

$\begin{aligned} P & = \text{3,000}\\ i &=\text{8.8} \% = \text{0.088}\\ n & = \text{6}\\ A & = \text{ ?} \end{aligned}$

This is a compound interest problem:

$\begin{aligned} \quad A = P(1+i)^n \end{aligned}$

Solve the first part of the problem. Remember that we must multiply the number of years by $\text{2}$ and divide the interest rate by $\text{2}$ because the interest is compounded every six months.

$\begin{aligned} & \\ A & = \text{3,000.0} \left( 1 + \frac{\text{0.088}} {\text{2}} \right) ^ {(\text{6} \times \text{2})}\\ A & = \text{3,000.0}(\text{1.044}...)^{\text{12}}\\ & = \text{3,000.0} (\text{1.676}...)\\ & = \text{N}\llap{=}\,\text{5,029.53} \end{aligned}$

This is the amount after $\text{6}$ years.

STEP: <no title>
[−3 points ⇒ 0 / 6 points left]

Determine what has been provided and what is required for the next $\text{4}$ years. All of the money from the above calculation remains in the account - it is the starting point for the last $\text{4}$ years.

$\begin{aligned} P & = \text{N}\llap{=}\,\text{5,029.53}\\ i &=\text{10.1} \% = \text{0.101}\\ n & = \text{4}\\ A & = \text{ ?} \end{aligned}$

Now solve the second part of the problem, the last $\text{4}$ years (again, we must adjust the interest rate and the the number of years by $\text{2}$ .):

$\begin{aligned} A & = \text{5,029.53} \left( 1 + \frac{\text{0.101}} {\text{2}} \right) ^ {(\text{4} \times \text{2})}\\ & = \text{5,029.53}(\text{1.050}...)^\text{8}\\ & = \text{5,029.53} (\text{1.483}...)\\ & = \text{N}\llap{=}\,\text{7,459.26} \end{aligned}$

After $\text{10}$ years, Emma will have $\text{N}\llap{=}\,\text{7,459.26}$ in her account.

Submit your answer as:
What is the total amount of interest the bank pays into her account?

Answer: Interest earned = R
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The amount of interest is what the bank paid into the account on top of what Emma put into the account at the beginning.

STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

Use the answer from the work above to calculate how much interest the bank paid.

$\begin{aligned} \text{Interest earned } & = \text{N}\llap{=}\,\text{7,459.26} - \text{N}\llap{=}\,\text{3,000.00} \\ & = \text{N}\llap{=}\,\text{4,459.26} \end{aligned}$

The amount of interest the bank pays is $\text{N}\llap{=}\,\text{4,459.26}$ .

Submit your answer as:

ID is: 1521 Seed is: 6179

Timeline questions: changing interest rates

Eniola goes to Save Here Bank and deposits $\text{N}\llap{=}\,\text{3,000.00}$ into an account which pays $\text{11.6}\%$ p.a. compounded every six months. After $\text{6}$ years the bank changes the interest rate to $\text{10.2}\%$ p.a., still compounded every six months.

How much money will Eniola have in her account $\text{8}$ years after the original investment?

Answer: Total amount = $\text{N}\llap{=}$
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

This question is a timeline question in which the interest rate changes. When does it change? How many times does it change? You need to break the question into separate pieces between each of the interest rate changes, and work through each of those pieces to get the final answer.

STEP: <no title>
[−3 points ⇒ 3 / 6 points left]

This question describes a $\text{8}$ year period which is broken into two sections because of the change in the interest rate: the first section lasts $\text{6}$ years (while the interest rate is $\text{11.6}\%$ ); the second section lasts $\text{2}$ years (when the rate is $\text{10.2}\%$ ).

Determine what has been provided and what is required for the first $\text{6}$ years:

$\begin{aligned} P & = \text{3,000}\\ i &=\text{11.6} \% = \text{0.116}\\ n & = \text{6}\\ A & = \text{ ?} \end{aligned}$

This is a compound interest problem:

$\begin{aligned} \quad A = P(1+i)^n \end{aligned}$

Solve the first part of the problem. Remember that we must multiply the number of years by $\text{2}$ and divide the interest rate by $\text{2}$ because the interest is compounded every six months.

$\begin{aligned} & \\ A & = \text{3,000.0} \left( 1 + \frac{\text{0.116}} {\text{2}} \right) ^ {(\text{6} \times \text{2})}\\ A & = \text{3,000.0}(\text{1.058}...)^{\text{12}}\\ & = \text{3,000.0} (\text{1.967}...)\\ & = \text{N}\llap{=}\,\text{5,901.32} \end{aligned}$

This is the amount after $\text{6}$ years.

STEP: <no title>
[−3 points ⇒ 0 / 6 points left]

Determine what has been provided and what is required for the next $\text{2}$ years. All of the money from the above calculation remains in the account - it is the starting point for the last $\text{2}$ years.

$\begin{aligned} P & = \text{N}\llap{=}\,\text{5,901.32}\\ i &=\text{10.2} \% = \text{0.102}\\ n & = \text{2}\\ A & = \text{ ?} \end{aligned}$

Now solve the second part of the problem, the last $\text{2}$ years (again, we must adjust the interest rate and the the number of years by $\text{2}$ .):

$\begin{aligned} A & = \text{5,901.32} \left( 1 + \frac{\text{0.102}} {\text{2}} \right) ^ {(\text{2} \times \text{2})}\\ & = \text{5,901.32}(\text{1.051}...)^\text{4}\\ & = \text{5,901.32} (\text{1.220}...)\\ & = \text{N}\llap{=}\,\text{7,200.46} \end{aligned}$

After $\text{8}$ years, Eniola will have $\text{N}\llap{=}\,\text{7,200.46}$ in her account.

Submit your answer as:
What is the total amount of interest the bank pays into her account?

Answer: Interest earned = R
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The amount of interest is what the bank paid into the account on top of what Eniola put into the account at the beginning.

STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

Use the answer from the work above to calculate how much interest the bank paid.

$\begin{aligned} \text{Interest earned } & = \text{N}\llap{=}\,\text{7,200.46} - \text{N}\llap{=}\,\text{3,000.00} \\ & = \text{N}\llap{=}\,\text{4,200.46} \end{aligned}$

The amount of interest the bank pays is $\text{N}\llap{=}\,\text{4,200.46}$ .

Submit your answer as:

ID is: 1521 Seed is: 3973

Timeline questions: changing interest rates

Gina goes to Save Here Bank and deposits $\text{N}\llap{=}\,\text{3,500.00}$ into an account which pays $\text{10.4}\%$ p.a. compounded semi-annually. After $\text{4}$ years the bank changes the interest rate to $\text{10.0}\%$ p.a., still compounded semi-annually.

How much money will Gina have in her account $\text{11}$ years after the original investment?

Answer: Total amount = $\text{N}\llap{=}$
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

This question is a timeline question in which the interest rate changes. When does it change? How many times does it change? You need to break the question into separate pieces between each of the interest rate changes, and work through each of those pieces to get the final answer.

STEP: <no title>
[−3 points ⇒ 3 / 6 points left]

This question describes a $\text{11}$ year period which is broken into two sections because of the change in the interest rate: the first section lasts $\text{4}$ years (while the interest rate is $\text{10.4}\%$ ); the second section lasts $\text{7}$ years (when the rate is $\text{10.0}\%$ ).

Determine what has been provided and what is required for the first $\text{4}$ years:

$\begin{aligned} P & = \text{3,500}\\ i &=\text{10.4} \% = \text{0.104}\\ n & = \text{4}\\ A & = \text{ ?} \end{aligned}$

This is a compound interest problem:

$\begin{aligned} \quad A = P(1+i)^n \end{aligned}$

Solve the first part of the problem. Remember that we must multiply the number of years by $\text{2}$ and divide the interest rate by $\text{2}$ because the interest is compounded semi-annually.

$\begin{aligned} & \\ A & = \text{3,500.0} \left( 1 + \frac{\text{0.104}} {\text{2}} \right) ^ {(\text{4} \times \text{2})}\\ A & = \text{3,500.0}(\text{1.052}...)^{\text{8}}\\ & = \text{3,500.0} (\text{1.500}...)\\ & = \text{N}\llap{=}\,\text{5,250.42} \end{aligned}$

This is the amount after $\text{4}$ years.

STEP: <no title>
[−3 points ⇒ 0 / 6 points left]

Determine what has been provided and what is required for the next $\text{7}$ years. All of the money from the above calculation remains in the account - it is the starting point for the last $\text{7}$ years.

$\begin{aligned} P & = \text{N}\llap{=}\,\text{5,250.42}\\ i &=\text{10.0} \% = \text{0.1}\\ n & = \text{7}\\ A & = \text{ ?} \end{aligned}$

Now solve the second part of the problem, the last $\text{7}$ years (again, we must adjust the interest rate and the the number of years by $\text{2}$ .):

$\begin{aligned} A & = \text{5,250.42} \left( 1 + \frac{\text{0.1}} {\text{2}} \right) ^ {(\text{7} \times \text{2})}\\ & = \text{5,250.42}(\text{1.050}...)^\text{14}\\ & = \text{5,250.42} (\text{1.979}...)\\ & = \text{N}\llap{=}\,\text{10,395.47} \end{aligned}$

After $\text{11}$ years, Gina will have $\text{N}\llap{=}\,\text{10,395.47}$ in her account.

Submit your answer as:
What is the total amount of interest the bank pays into her account?

Answer: Interest earned = R
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The amount of interest is what the bank paid into the account on top of what Gina put into the account at the beginning.

STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

Use the answer from the work above to calculate how much interest the bank paid.

$\begin{aligned} \text{Interest earned } & = \text{N}\llap{=}\,\text{10,395.47} - \text{N}\llap{=}\,\text{3,500.00} \\ & = \text{N}\llap{=}\,\text{6,895.47} \end{aligned}$

The amount of interest the bank pays is $\text{N}\llap{=}\,\text{6,895.47}$ .

Submit your answer as:

ID is: 310 Seed is: 3707

Compound interest investment

An amount of $\text{N}\llap{=}\,\text{2,520}$ is invested in a savings account which pays a compound interest rate of $\text{9.2}\%$ p.a.

Calculate the balance accumulated by the end of $\text{5}$ years. As usual with financial calculations, round your answer to two decimal places, but do not round off until you have reached the solution.

Answer: The final balance is R .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

This question is about compound interest. Start by writing down the compound interest formula. Then figure out what values go into the formula.

STEP: Write down the information in the question
[−1 point ⇒ 1 / 2 points left]

Read the question carefully and write down the given information:

$A = ?$
$P = \text{2,520}$
$i = \frac{\text{9.2}}{100}$
$n = \text{5}$

STEP: Use the compound interest formula to calculate the answer
[−1 point ⇒ 0 / 2 points left]

The accumulated amount is:

\begin{aligned} A & = P (1 + i)^{n} \\ = 2,520 {(1 + \frac{9.2}{100})}^{5} \\ = 2,520 {(1.092)}^{5} \\ = N = 3,913.04 \end{aligned}

$\begin{aligned} A &= P (1+i)^n\\ &= \text{2,520} \left(1 + \frac{\text{9.2}}{\text{100}}\right)^{\text{5}} \\ &= \text{2,520} \left( \text{1.092} \right)^{\text{5}} \\ &= \text{N}\llap{=}\,\text{3,913.04} \end{aligned}$

If you have an answer close to this, it is likely that you rounded off during the calculation. However, you should not round off until you reach the end of the calculation - let your calculator hold all of the decimal digits that come as you do the calculation until the end.

The balance after years is $\text{N}\llap{=}\,\text{3,913.04}$ .

Submit your answer as:

ID is: 310 Seed is: 1854

Compound interest investment

An amount of $\text{N}\llap{=}\,\text{3,000}$ is invested in a savings account which pays a compound interest rate of $\text{7.5}\%$ p.a.

Calculate the balance accumulated by the end of $\text{6}$ years. As usual with financial calculations, round your answer to two decimal places, but do not round off until you have reached the solution.

Answer: The final balance is R .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

This question is about compound interest. Start by writing down the compound interest formula. Then figure out what values go into the formula.

STEP: Write down the information in the question
[−1 point ⇒ 1 / 2 points left]

Read the question carefully and write down the given information:

$A = ?$
$P = \text{3,000}$
$i = \frac{\text{7.5}}{100}$
$n = \text{6}$

STEP: Use the compound interest formula to calculate the answer
[−1 point ⇒ 0 / 2 points left]

The accumulated amount is:

\begin{aligned} A & = P (1 + i)^{n} \\ = 3,000 {(1 + \frac{7.5}{100})}^{6} \\ = 3,000 {(1.075)}^{6} \\ = N = 4,629.90 \end{aligned}

$\begin{aligned} A &= P (1+i)^n\\ &= \text{3,000} \left(1 + \frac{\text{7.5}}{\text{100}}\right)^{\text{6}} \\ &= \text{3,000} \left( \text{1.075} \right)^{\text{6}} \\ &= \text{N}\llap{=}\,\text{4,629.90} \end{aligned}$

The balance after years is $\text{N}\llap{=}\,\text{4,629.90}$ .

Submit your answer as:

ID is: 310 Seed is: 9208

Compound interest investment

An amount of $\text{N}\llap{=}\,\text{6,120}$ is invested in a savings account which pays a compound interest rate of $\text{9.3}\%$ p.a.

Calculate the balance accumulated by the end of $\text{3}$ years. As usual with financial calculations, round your answer to two decimal places, but do not round off until you have reached the solution.

Answer: The final balance is R .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

This question is about compound interest. Start by writing down the compound interest formula. Then figure out what values go into the formula.

STEP: Write down the information in the question
[−1 point ⇒ 1 / 2 points left]

Read the question carefully and write down the given information:

$A = ?$
$P = \text{6,120}$
$i = \frac{\text{9.3}}{100}$
$n = \text{3}$

STEP: Use the compound interest formula to calculate the answer
[−1 point ⇒ 0 / 2 points left]

The accumulated amount is:

\begin{aligned} A & = P (1 + i)^{n} \\ = 6,120 {(1 + \frac{9.3}{100})}^{3} \\ = 6,120 {(1.093)}^{3} \\ = N = 7,991.20 \end{aligned}

$\begin{aligned} A &= P (1+i)^n\\ &= \text{6,120} \left(1 + \frac{\text{9.3}}{\text{100}}\right)^{\text{3}} \\ &= \text{6,120} \left( \text{1.093} \right)^{\text{3}} \\ &= \text{N}\llap{=}\,\text{7,991.20} \end{aligned}$

The balance after years is $\text{N}\llap{=}\,\text{7,991.20}$ .

Submit your answer as:

ID is: 1520 Seed is: 2594

Timelines: population growth

Farmer Methuli starts working with a beehive in his orchard. The beehive contains $\text{59,000}$ bees. The number of bees in the hive grows by $\text{2.51}\%$ each year. After $\text{7}$ years farmer Methuli gives $\text{9,500}$ of his bees to another farmer who is starting another hive. Both hives continue to grow at the same rate ( $\text{2.51}\%$ every year).

How many bees will be in farmer Methuli's hive $\text{5}$ years after he gave some of his bees away?

INSTRUCTION: Round your answer to the nearest whole number.
Answer:
There will be bees in Methuli's hive (rounded to the nearest whole number).

one-of
type(numeric.abserror(1.0))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Populations grow following the same compounding pattern as compound interest in a savings account. Think about what equation you should use for this question.

STEP: Calculate the number of bees after $\text{7}$ years
[−2 points ⇒ 2 / 4 points left]

Populations grow like compound interest, so we should use this formula: $\; A = P(1+i)^n$ .

In this case we will use the equation as follows:
- $P$ is the number of bees at the beginning $= \text{59,000}$
- $i$ is the rate of growth of th hive $= \text{2.51}\% = \text{0.0251}$
- $n$ is the number of years $= \text{7}$
- $A = ?$
Solve the first part of the problem:

$\begin{aligned} A & = \text{59,000} \left( 1 + \text{0.0251}\right) ^ \text{7}\\ & = \text{70,180.368...} \\ & \approx \text{70,180} \end{aligned}$

TIP: We cannot know the exact number of bees in the hive; this formula is just an estimate. To get the most accurate estimate, you should keep this number saved in your calculator and use it in the following calculations.

So, there are approximately $\text{70,180}$ bees in the hive after $\text{7}$ years.

STEP: Calculate the number of bees after $\text{12}$ years
[−2 points ⇒ 0 / 4 points left]

At this point the farmer takes away (subtracts) $\text{9,500}$ bees from his hive. You must subtract now because Methuli is taking bees out of the hive.

$\begin{aligned} P & = \text{70,180} - \text{9,500} = \text{60,680} \\ i &=\text{2.51}\% = \text{0.0251}\\ n & = \text{5}\\ A & = \text{ ?} \end{aligned}$

Now calculate how many bees there will be after the last $\text{5}$ years.

$\begin{aligned} A & = \text{60,680} \left( 1 + \text{0.0251} \right) ^ \text{5} \\ & = \text{68,687.346...} \\ & \approx \text{68,687} \end{aligned}$

So, after $\text{12}$ years, farmer Methuli will have $\text{68,687}$ bees in his beehive.

NOTE:
You could have also calculated this in one step, by moving all of the quantities to the end of the time period:

$\begin{aligned} A & = \text{59,000} \left( 1 + \text{0.0251}\right) ^ \text{12} - \text{9,500} \left( 1 + \text{0.0251} \right) ^ \text{5}\\ & = \text{68,687.764...} \\ & \approx \text{68,688} \end{aligned}$

This sometimes gives a different answer than what you get when you round off at each step. But, it should be marked as correct because it is actually more accurate.

Submit your answer as:
There were $\text{9,500}$ in the second hive when it started. How many bees will be in the second hive after $\text{5}$ years?

INSTRUCTION: Round your answer to the nearest whole number.

Answer:
There will be bees in the second hive (rounded to the nearest whole number).

numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You must figure out how many years the second bee hive has to grow (it is not the same as the first hive).

STEP: Calculate the number of bees in the second hive after $\text{5}$ years
[−2 points ⇒ 0 / 2 points left]

For this question, remember that the second hive starts with $\text{9,500}$ bees, but it only starts $\text{7}$ years after the first hive. Therefore the second hive only grows for $\text{5}$ years.

$\begin{aligned} P & = \text{9,500} \\ i &=\text{2.51}\% = \text{0.0251}\\ n & = \text{5}\\ A & = \text{ ?} \end{aligned}$

$\begin{aligned} A & = \text{9,500} \left( 1 + \text{0.0251} \right) ^ \text{5} \\ & = \text{10,753.622...} \\ & \approx \text{10,754} \end{aligned}$

So, after $\text{5}$ years, the second hive had approximately $\text{10,754}$ bees in it.

Submit your answer as:
A single bee can make about $\text{0.56}\ \text{grams}$ of honey in a summer. If $\text{90}\%$ of the bees in the colony produce honey, how much honey did the two hives produce in the $12^{\text{th}}$ year after farmer Methuli started working with bees?

INSTRUCTION: Write your answer in $\text{kilograms}$ and then round it to two decimal places.

Answer:
The amount of honey produced is $\text{kg}$ .

one-of
type(numeric.abserror(0.01))
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

How many bees are there in both hives after $\text{12}$ years?

STEP: Calculate the number of bees that produce honey
[−1 point ⇒ 1 / 2 points left]

$\text{90}\%$ of the bees are producing honey (yum!). To figure out how many bees are producing honey in the two hives, we must calculate $\text{90}\%$ ( $\text{0.90}$ ) of the total number of bees:

$\begin{aligned} & \text{0.90} \times (\text{68,687} + \text{10,754}) \\ & = \text{71,496.9} \\ & \approx \text{71,497} \end{aligned}$

TIP: We cannot know the exact number of bees that produce honey; this formula is just an estimate. To get the most accurate estimate, you should keep this number saved in your calculator and use it in the following calculations.

So there are about $\text{71,497}$ bees making honey.

STEP: Calculate how much honey these bees make, by multiplying
[−0 points ⇒ 1 / 2 points left]

Each of the $\text{71,497}$ bees produces $\text{0.56}$ grams of honey. To find the total amount of honey, we just multiply:

$\begin{aligned} & \text{0.56}\: \times \text{71,497} \\ & = \text{40,038.32} \end{aligned}$

STEP: Convert grams to kilograms
[−1 point ⇒ 0 / 2 points left]

But we aren't done yet because the question asks us to write the answer in kilograms. Every $\text{kg}$ is equal to $\text{1,000} \ \text{g}$ .

The amount of honey (in $\text{kg}$ ) is:

$\begin{aligned} & \frac{\text{40,038.32}}{1000} \\ & = \text{40.03832} \\ & \approx \text{40.04}\ \text{kg} \end{aligned}$

The two hives produce approximately $\text{40.04}\ \text{kg}$ of honey.

Submit your answer as:

ID is: 1520 Seed is: 7162

Timelines: population growth

Farmer Abidemi starts working with a beehive in his orchard. The beehive contains $\text{56,000}$ bees. The number of bees in the hive grows by $\text{3.13}\%$ each year. After $\text{6}$ years farmer Abidemi gives $\text{9,500}$ of his bees to another farmer who is starting another hive. Both hives continue to grow at the same rate ( $\text{3.13}\%$ every year).

How many bees will be in farmer Abidemi's hive $\text{4}$ years after he gave some of his bees away?

INSTRUCTION: Round your answer to the nearest whole number.
Answer:
There will be bees in Abidemi's hive (rounded to the nearest whole number).

one-of
type(numeric.abserror(1.0))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Populations grow following the same compounding pattern as compound interest in a savings account. Think about what equation you should use for this question.

STEP: Calculate the number of bees after $\text{6}$ years
[−2 points ⇒ 2 / 4 points left]

Populations grow like compound interest, so we should use this formula: $\; A = P(1+i)^n$ .

In this case we will use the equation as follows:
- $P$ is the number of bees at the beginning $= \text{56,000}$
- $i$ is the rate of growth of th hive $= \text{3.13}\% = \text{0.0313}$
- $n$ is the number of years $= \text{6}$
- $A = ?$
Solve the first part of the problem:

$\begin{aligned} A & = \text{56,000} \left( 1 + \text{0.0313}\right) ^ \text{6}\\ & = \text{67,374.899...} \\ & \approx \text{67,375} \end{aligned}$

TIP: We cannot know the exact number of bees in the hive; this formula is just an estimate. To get the most accurate estimate, you should keep this number saved in your calculator and use it in the following calculations.

So, there are approximately $\text{67,375}$ bees in the hive after $\text{6}$ years.

STEP: Calculate the number of bees after $\text{10}$ years
[−2 points ⇒ 0 / 4 points left]

At this point the farmer takes away (subtracts) $\text{9,500}$ bees from his hive. You must subtract now because Abidemi is taking bees out of the hive.

$\begin{aligned} P & = \text{67,375} - \text{9,500} = \text{57,875} \\ i &=\text{3.13}\% = \text{0.0313}\\ n & = \text{4}\\ A & = \text{ ?} \end{aligned}$

Now calculate how many bees there will be after the last $\text{4}$ years.

$\begin{aligned} A & = \text{57,875} \left( 1 + \text{0.0313} \right) ^ \text{4} \\ & = \text{65,468.301...} \\ & \approx \text{65,468} \end{aligned}$

So, after $\text{10}$ years, farmer Abidemi will have $\text{65,468}$ bees in his beehive.

NOTE:
You could have also calculated this in one step, by moving all of the quantities to the end of the time period:

$\begin{aligned} A & = \text{56,000} \left( 1 + \text{0.0313}\right) ^ \text{10} - \text{9,500} \left( 1 + \text{0.0313} \right) ^ \text{4}\\ & = \text{65,468.188...} \\ & \approx \text{65,468} \end{aligned}$

This sometimes gives a different answer than what you get when you round off at each step. But, it should be marked as correct because it is actually more accurate.

Submit your answer as:
There were $\text{9,500}$ in the second hive when it started. How many bees will be in the second hive after $\text{4}$ years?

INSTRUCTION: Round your answer to the nearest whole number.

Answer:
There will be bees in the second hive (rounded to the nearest whole number).

numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You must figure out how many years the second bee hive has to grow (it is not the same as the first hive).

STEP: Calculate the number of bees in the second hive after $\text{4}$ years
[−2 points ⇒ 0 / 2 points left]

For this question, remember that the second hive starts with $\text{9,500}$ bees, but it only starts $\text{6}$ years after the first hive. Therefore the second hive only grows for $\text{4}$ years.

$\begin{aligned} P & = \text{9,500} \\ i &=\text{3.13}\% = \text{0.0313}\\ n & = \text{4}\\ A & = \text{ ?} \end{aligned}$

$\begin{aligned} A & = \text{9,500} \left( 1 + \text{0.0313} \right) ^ \text{4} \\ & = \text{10,746.416...} \\ & \approx \text{10,746} \end{aligned}$

So, after $\text{4}$ years, the second hive had approximately $\text{10,746}$ bees in it.

Submit your answer as:
A single bee can make about $\text{0.56}\ \text{grams}$ of honey in a summer. If $\text{90}\%$ of the bees in the colony produce honey, how much honey did the two hives produce in the tenth year after farmer Abidemi started working with bees?

INSTRUCTION: Write your answer in $\text{kilograms}$ and then round it to two decimal places.

Answer:
The amount of honey produced is $\text{kg}$ .

one-of
type(numeric.abserror(0.01))
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

How many bees are there in both hives after $\text{10}$ years?

STEP: Calculate the number of bees that produce honey
[−1 point ⇒ 1 / 2 points left]

$\text{90}\%$ of the bees are producing honey (yum!). To figure out how many bees are producing honey in the two hives, we must calculate $\text{90}\%$ ( $\text{0.90}$ ) of the total number of bees:

$\begin{aligned} & \text{0.90} \times (\text{65,468} + \text{10,746}) \\ & = \text{68,592.6} \\ & \approx \text{68,593} \end{aligned}$

TIP: We cannot know the exact number of bees that produce honey; this formula is just an estimate. To get the most accurate estimate, you should keep this number saved in your calculator and use it in the following calculations.

So there are about $\text{68,593}$ bees making honey.

STEP: Calculate how much honey these bees make, by multiplying
[−0 points ⇒ 1 / 2 points left]

Each of the $\text{68,593}$ bees produces $\text{0.56}$ grams of honey. To find the total amount of honey, we just multiply:

$\begin{aligned} & \text{0.56}\: \times \text{68,593} \\ & = \text{38,412.08} \end{aligned}$

STEP: Convert grams to kilograms
[−1 point ⇒ 0 / 2 points left]

But we aren't done yet because the question asks us to write the answer in kilograms. Every $\text{kg}$ is equal to $\text{1,000} \ \text{g}$ .

The amount of honey (in $\text{kg}$ ) is:

$\begin{aligned} & \frac{\text{38,412.08}}{1000} \\ & = \text{38.41208} \\ & \approx \text{38.41}\ \text{kg} \end{aligned}$

The two hives produce approximately $\text{38.41}\ \text{kg}$ of honey.

Submit your answer as:

ID is: 1520 Seed is: 4109

Timelines: population growth

Farmer Adedamola starts working with a beehive in his orchard. The beehive contains $\text{63,000}$ bees. The number of bees in the hive grows by $\text{2.42}\%$ each year. After $\text{6}$ years farmer Adedamola gives $\text{11,000}$ of his bees to another farmer who is starting another hive. Both hives continue to grow at the same rate ( $\text{2.42}\%$ every year).

How many bees will be in farmer Adedamola's hive $\text{4}$ years after he gave some of his bees away?

INSTRUCTION: Round your answer to the nearest whole number.
Answer:
There will be bees in Adedamola's hive (rounded to the nearest whole number).

one-of
type(numeric.abserror(1.0))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Populations grow following the same compounding pattern as compound interest in a savings account. Think about what equation you should use for this question.

STEP: Calculate the number of bees after $\text{6}$ years
[−2 points ⇒ 2 / 4 points left]

Populations grow like compound interest, so we should use this formula: $\; A = P(1+i)^n$ .

In this case we will use the equation as follows:
- $P$ is the number of bees at the beginning $= \text{63,000}$
- $i$ is the rate of growth of th hive $= \text{2.42}\% = \text{0.0242}$
- $n$ is the number of years $= \text{6}$
- $A = ?$
Solve the first part of the problem:

$\begin{aligned} A & = \text{63,000} \left( 1 + \text{0.0242}\right) ^ \text{6}\\ & = \text{72,719.214...} \\ & \approx \text{72,719} \end{aligned}$

TIP: We cannot know the exact number of bees in the hive; this formula is just an estimate. To get the most accurate estimate, you should keep this number saved in your calculator and use it in the following calculations.

So, there are approximately $\text{72,719}$ bees in the hive after $\text{6}$ years.

STEP: Calculate the number of bees after $\text{10}$ years
[−2 points ⇒ 0 / 4 points left]

At this point the farmer takes away (subtracts) $\text{11,000}$ bees from his hive. You must subtract now because Adedamola is taking bees out of the hive.

$\begin{aligned} P & = \text{72,719} - \text{11,000} = \text{61,719} \\ i &=\text{2.42}\% = \text{0.0242}\\ n & = \text{4}\\ A & = \text{ ?} \end{aligned}$

Now calculate how many bees there will be after the last $\text{4}$ years.

$\begin{aligned} A & = \text{61,719} \left( 1 + \text{0.0242} \right) ^ \text{4} \\ & = \text{67,913.789...} \\ & \approx \text{67,914} \end{aligned}$

So, after $\text{10}$ years, farmer Adedamola will have $\text{67,914}$ bees in his beehive.

NOTE:
You could have also calculated this in one step, by moving all of the quantities to the end of the time period:

$\begin{aligned} A & = \text{63,000} \left( 1 + \text{0.0242}\right) ^ \text{10} - \text{11,000} \left( 1 + \text{0.0242} \right) ^ \text{4}\\ & = \text{67,914.025...} \\ & \approx \text{67,914} \end{aligned}$

This sometimes gives a different answer than what you get when you round off at each step. But, it should be marked as correct because it is actually more accurate.

Submit your answer as:
There were $\text{11,000}$ in the second hive when it started. How many bees will be in the second hive after $\text{4}$ years?

INSTRUCTION: Round your answer to the nearest whole number.

Answer:
There will be bees in the second hive (rounded to the nearest whole number).

numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You must figure out how many years the second bee hive has to grow (it is not the same as the first hive).

STEP: Calculate the number of bees in the second hive after $\text{4}$ years
[−2 points ⇒ 0 / 2 points left]

For this question, remember that the second hive starts with $\text{11,000}$ bees, but it only starts $\text{6}$ years after the first hive. Therefore the second hive only grows for $\text{4}$ years.

$\begin{aligned} P & = \text{11,000} \\ i &=\text{2.42}\% = \text{0.0242}\\ n & = \text{4}\\ A & = \text{ ?} \end{aligned}$

$\begin{aligned} A & = \text{11,000} \left( 1 + \text{0.0242} \right) ^ \text{4} \\ & = \text{12,104.079...} \\ & \approx \text{12,104} \end{aligned}$

So, after $\text{4}$ years, the second hive had approximately $\text{12,104}$ bees in it.

Submit your answer as:
A single bee can make about $\text{0.56}\ \text{grams}$ of honey in a summer. If $\text{90}\%$ of the bees in the colony produce honey, how much honey did the two hives produce in the tenth year after farmer Adedamola started working with bees?

INSTRUCTION: Write your answer in $\text{kilograms}$ and then round it to two decimal places.

Answer:
The amount of honey produced is $\text{kg}$ .

one-of
type(numeric.abserror(0.01))
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

How many bees are there in both hives after $\text{10}$ years?

STEP: Calculate the number of bees that produce honey
[−1 point ⇒ 1 / 2 points left]

$\text{90}\%$ of the bees are producing honey (yum!). To figure out how many bees are producing honey in the two hives, we must calculate $\text{90}\%$ ( $\text{0.90}$ ) of the total number of bees:

$\begin{aligned} & \text{0.90} \times (\text{67,914} + \text{12,104}) \\ & = \text{72,016.2} \\ & \approx \text{72,016} \end{aligned}$

TIP: We cannot know the exact number of bees that produce honey; this formula is just an estimate. To get the most accurate estimate, you should keep this number saved in your calculator and use it in the following calculations.

So there are about $\text{72,016}$ bees making honey.

STEP: Calculate how much honey these bees make, by multiplying
[−0 points ⇒ 1 / 2 points left]

Each of the $\text{72,016}$ bees produces $\text{0.56}$ grams of honey. To find the total amount of honey, we just multiply:

$\begin{aligned} & \text{0.56}\: \times \text{72,016} \\ & = \text{40,328.96} \end{aligned}$

STEP: Convert grams to kilograms
[−1 point ⇒ 0 / 2 points left]

But we aren't done yet because the question asks us to write the answer in kilograms. Every $\text{kg}$ is equal to $\text{1,000} \ \text{g}$ .

The amount of honey (in $\text{kg}$ ) is:

$\begin{aligned} & \frac{\text{40,328.96}}{1000} \\ & = \text{40.32896} \\ & \approx \text{40.33}\ \text{kg} \end{aligned}$

The two hives produce approximately $\text{40.33}\ \text{kg}$ of honey.

Submit your answer as:

2. Depreciation

ID is: 1534 Seed is: 2338

Increasing values: finding i

After $\text{8}$ years an investment quadruples (gets 4 times bigger) in value. At what annual percentage rate was interest compounded? Give your answer correct to two decimal places.

Answer: % per annum.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Read the question carefully and write down the given information.

STEP: <no title>
[−0 points ⇒ 4 / 4 points left]

Write down the given information. You don't know the value of P in this question, but you do know that the final value of the account is 4 times as much as at the beginning:

$A = \text{4} \times P$
$P = P$
$n = \text{8}$
$i = ?$

STEP: <no title>
[−0 points ⇒ 4 / 4 points left]

The question says, "at what annual percentage rate was interest compounded;" that tells you that the account follows the compound interest formula.

\begin{matrix} A = P (1 + i)^{n} \end{matrix}

$\begin{array}{c}A = P(1+i)^{n}\end{array}$

STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Now substitute in the values you know. At this point, you still don't know the value of P, but watch what happens!

\begin{aligned} 4 P & = P {(1 + \frac{i}{100})}^{8} \\ 4 & = {(1 + \frac{i}{100})}^{8} \leftarrow_{each other!}^{The P's cancel} \end{aligned}

$\begin{aligned} \text{4}P& = P\left(\text{1}+\frac{i}{\text{100}}\right)^{\text{8}} \\ \text{4}& = \left( \text{1}+\frac{i}{\text{100}} \right) ^{\text{8}} \quad \leftarrow \quad ^{\text{The P's cancel}}_{\text{each other!}} \end{aligned}$

STEP: <no title>
[−2 points ⇒ 1 / 4 points left]

Now you must solve for $i$ . The first step is to use what you know about exponents - how can you cancel the power of $\text{8}$ in the equation? You must use the eighth root, as shown below:

$\begin{aligned} \sqrt[8]{(\text{4})} & = \sqrt[8] { \left( \text{1}+\frac{i}{\text{100}} \right) ^{\text{8}} } \\ 1,1892\ldots & = \text{1}+\frac{i}{\text{100}} \\ 0,1892\ldots & = \frac{i}{\text{100}} \\ \left( 0,1892\ldots \right) \times \text{100} & = i \\ i & = 18,9207\ldots \end{aligned}$

STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

Rounding to two decimal places as required by the question, you get the final answer:

\begin{matrix} Therefore i = 18.92 % \end{matrix}

$\begin{array}{c}\text{Therefore } i = \text{18.92}\%\end{array}$

Submit your answer as:

ID is: 1534 Seed is: 4542

Increasing values: finding i

After $\text{9}$ years an investment doubles in value. At what annual percentage rate was interest compounded? Give your answer correct to two decimal places.

Answer: % per annum.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Read the question carefully and write down the given information.

STEP: <no title>
[−0 points ⇒ 4 / 4 points left]

Write down the given information. You don't know the value of P in this question, but you do know that the final value of the account is 2 times as much as at the beginning:

$A = \text{2} \times P$
$P = P$
$n = \text{9}$
$i = ?$

STEP: <no title>
[−0 points ⇒ 4 / 4 points left]

The question says, "at what annual percentage rate was interest compounded;" that tells you that the account follows the compound interest formula.

\begin{matrix} A = P (1 + i)^{n} \end{matrix}

$\begin{array}{c}A = P(1+i)^{n}\end{array}$

STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Now substitute in the values you know. At this point, you still don't know the value of P, but watch what happens!

\begin{aligned} 2 P & = P {(1 + \frac{i}{100})}^{9} \\ 2 & = {(1 + \frac{i}{100})}^{9} \leftarrow_{each other!}^{The P's cancel} \end{aligned}

$\begin{aligned} \text{2}P& = P\left(\text{1}+\frac{i}{\text{100}}\right)^{\text{9}} \\ \text{2}& = \left( \text{1}+\frac{i}{\text{100}} \right) ^{\text{9}} \quad \leftarrow \quad ^{\text{The P's cancel}}_{\text{each other!}} \end{aligned}$

STEP: <no title>
[−2 points ⇒ 1 / 4 points left]

Now you must solve for $i$ . The first step is to use what you know about exponents - how can you cancel the power of $\text{9}$ in the equation? You must use the ninth root, as shown below:

$\begin{aligned} \sqrt[9]{(\text{2})} & = \sqrt[9] { \left( \text{1}+\frac{i}{\text{100}} \right) ^{\text{9}} } \\ 1,0800\ldots & = \text{1}+\frac{i}{\text{100}} \\ 0,0800\ldots & = \frac{i}{\text{100}} \\ \left( 0,0800\ldots \right) \times \text{100} & = i \\ i & = 8,0059\ldots \end{aligned}$

STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

Rounding to two decimal places as required by the question, you get the final answer:

\begin{matrix} Therefore i = 8.01 % \end{matrix}

$\begin{array}{c}\text{Therefore } i = \text{8.01}\%\end{array}$

Submit your answer as:

ID is: 1534 Seed is: 3467

Increasing values: finding i

After $\text{4}$ years an investment doubles in value. At what annual percentage rate was interest compounded? Give your answer correct to two decimal places.

Answer: % per annum.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Read the question carefully and write down the given information.

STEP: <no title>
[−0 points ⇒ 4 / 4 points left]

Write down the given information. You don't know the value of P in this question, but you do know that the final value of the account is 2 times as much as at the beginning:

$A = \text{2} \times P$
$P = P$
$n = \text{4}$
$i = ?$

STEP: <no title>
[−0 points ⇒ 4 / 4 points left]

The question says, "at what annual percentage rate was interest compounded;" that tells you that the account follows the compound interest formula.

\begin{matrix} A = P (1 + i)^{n} \end{matrix}

$\begin{array}{c}A = P(1+i)^{n}\end{array}$

STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Now substitute in the values you know. At this point, you still don't know the value of P, but watch what happens!

\begin{aligned} 2 P & = P {(1 + \frac{i}{100})}^{4} \\ 2 & = {(1 + \frac{i}{100})}^{4} \leftarrow_{each other!}^{The P's cancel} \end{aligned}

$\begin{aligned} \text{2}P& = P\left(\text{1}+\frac{i}{\text{100}}\right)^{\text{4}} \\ \text{2}& = \left( \text{1}+\frac{i}{\text{100}} \right) ^{\text{4}} \quad \leftarrow \quad ^{\text{The P's cancel}}_{\text{each other!}} \end{aligned}$

STEP: <no title>
[−2 points ⇒ 1 / 4 points left]

Now you must solve for $i$ . The first step is to use what you know about exponents - how can you cancel the power of $\text{4}$ in the equation? You must use the fourth root, as shown below:

$\begin{aligned} \sqrt[4]{(\text{2})} & = \sqrt[4] { \left( \text{1}+\frac{i}{\text{100}} \right) ^{\text{4}} } \\ 1,1892\ldots & = \text{1}+\frac{i}{\text{100}} \\ 0,1892\ldots & = \frac{i}{\text{100}} \\ \left( 0,1892\ldots \right) \times \text{100} & = i \\ i & = 18,9207\ldots \end{aligned}$

STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

Rounding to two decimal places as required by the question, you get the final answer:

\begin{matrix} Therefore i = 18.92 % \end{matrix}

$\begin{array}{c}\text{Therefore } i = \text{18.92}\%\end{array}$

Submit your answer as:

ID is: 1528 Seed is: 4537

Simple depreciation: when an asset goes to zero value!

A small business buys a printer for $\text{N}\llap{=}\,\text{5,000}$ . For the tax return, the owner depreciates this asset over $\text{6}$ years on a straight-line basis. Each year the company's owner must fill in the value of the printer on this tax form. What amount will he put on his tax form after $\text{2}$ year(s)?

Answer:

R

$R$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

In this question, the value of the printer decreases each and every year until the value has become $\text{N}\llap{=}\,\text{0}$ after $\text{6}$ years. Start by calculating how much the value decreases for each year.

STEP: Calculate the yearly depreciation
[−1 point ⇒ 2 / 3 points left]

The owner of the business wants the printer to depreciate to $\text{N}\llap{=}\,\text{0}$ after $\text{6}$ years. Since he is using straight-line depreciation, the printer will decrease by the same amount every year. This must be the value of the printer divided by the number of years: $\text{5,000} \div \text{6} =$ $\text{N}\llap{=}\,\text{833.33}$ per year.

STEP: Calculate the total depreciation
[−1 point ⇒ 1 / 3 points left]

Now find out how much value the printer lost in $\text{2}$ year(s), as the question asks. It will be the value lost per year (from above) multiplied by the number of years:

$\textrm{value lost} = \text{2}(\text{833.33}) = \text{1,666.66}$

STEP: Calculate the value of the printer
[−1 point ⇒ 0 / 3 points left]

Therefore the value of the printer after $\text{2}$ year/s is:

$\text{5,000} - \text{1,666.66} = \text{N}\llap{=}\,\text{3,333.34}$

Submit your answer as:

ID is: 1528 Seed is: 7852

Simple depreciation: when an asset goes to zero value!

A small business buys a computer for $\text{N}\llap{=}\,\text{15,000}$ . For the tax return, the owner depreciates this asset over $\text{7}$ years using a straight-line depreciation method. Each year the company's owner must fill in the value of the computer on this tax form. What amount will he put on his tax form after $\text{5}$ year(s)?

Answer:

R

$R$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

In this question, the value of the computer decreases each and every year until the value has become $\text{N}\llap{=}\,\text{0}$ after $\text{7}$ years. Start by calculating how much the value decreases for each year.

STEP: Calculate the yearly depreciation
[−1 point ⇒ 2 / 3 points left]

The owner of the business wants the computer to depreciate to $\text{N}\llap{=}\,\text{0}$ after $\text{7}$ years. Since he is using straight-line depreciation, the computer will decrease by the same amount every year. This must be the value of the computer divided by the number of years: $\text{15,000} \div \text{7} =$ $\text{N}\llap{=}\,\text{2,142.86}$ per year.

STEP: Calculate the total depreciation
[−1 point ⇒ 1 / 3 points left]

Now find out how much value the computer lost in $\text{5}$ year(s), as the question asks. It will be the value lost per year (from above) multiplied by the number of years:

$\textrm{value lost} = \text{5}(\text{2,142.86}) = \text{10,714.3}$

STEP: Calculate the value of the computer
[−1 point ⇒ 0 / 3 points left]

Therefore the value of the computer after $\text{5}$ year/s is:

$\text{15,000} - \text{10,714.3} = \text{N}\llap{=}\,\text{4,285.70}$

Submit your answer as:

ID is: 1528 Seed is: 81

Simple depreciation: when an asset goes to zero value!

A small business buys a fridge for $\text{N}\llap{=}\,\text{7,000}$ . For the tax return, the owner depreciates this asset over $\text{9}$ years following the straight-line method. Each year the company's owner must fill in the value of the fridge on this tax form. What amount will he put on his tax form after $\text{3}$ year(s)?

Answer:

R

$R$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

In this question, the value of the fridge decreases each and every year until the value has become $\text{N}\llap{=}\,\text{0}$ after $\text{9}$ years. Start by calculating how much the value decreases for each year.

STEP: Calculate the yearly depreciation
[−1 point ⇒ 2 / 3 points left]

The owner of the business wants the fridge to depreciate to $\text{N}\llap{=}\,\text{0}$ after $\text{9}$ years. Since he is using straight-line depreciation, the fridge will decrease by the same amount every year. This must be the value of the fridge divided by the number of years: $\text{7,000} \div \text{9} =$ $\text{N}\llap{=}\,\text{777.78}$ per year.

STEP: Calculate the total depreciation
[−1 point ⇒ 1 / 3 points left]

Now find out how much value the fridge lost in $\text{3}$ year(s), as the question asks. It will be the value lost per year (from above) multiplied by the number of years:

$\textrm{value lost} = \text{3}(\text{777.78}) = \text{2,333.34}$

STEP: Calculate the value of the fridge
[−1 point ⇒ 0 / 3 points left]

Therefore the value of the fridge after $\text{3}$ year/s is:

$\text{7,000} - \text{2,333.34} = \text{N}\llap{=}\,\text{4,666.66}$

Submit your answer as:

ID is: 1524 Seed is: 7802

Compound interest: finding $i$

Xabanisile deposits $\text{N}\llap{=}\,\text{7,000}$ into a savings account which earns compound interest. The account grows to a value of $\text{N}\llap{=}\,\text{9,919.21}$ in $\text{5}$ years. Determine the value of $i$ , the rate of interest paid by the bank.

INSTRUCTION: Give your answer as a percentage rounded to two decimal places.

Answer: % per annum.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Does this account involve simple interest or compound interest? You need to know that so you know which formula to use.

STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Write down the given information.

$A = \text{N}\llap{=}\,\text{9,919.21}$
$P = \text{N}\llap{=}\,\text{7,000}$
$n = \text{5}$
$i = ?$

The question says, "earns compound interest," so we need to use the compound interest formula. Substitute in what you know!

\begin{aligned} A & = P (1 + i)^{n} \\ 9,919.21 & = 7,000 {(1 + i)}^{5} \end{aligned}

$\begin{aligned} A & = P(1+i)^{n} \\ \text{9,919.21} & = \text{7,000}\left( 1 + i \right)^{\text{5}} \end{aligned}$

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

Solve for $i$ . The first step is to divide by $\text{7,000}$ . Then we will deal with the exponent.

\begin{aligned} 9,919.21 & = 7,000 {(1 + i)}^{5} \\ \frac{9,919.21}{7,000} & = {(1 + i)}^{5} \\ 1, 41703 \dots & = {(1 + i)}^{5} \end{aligned}

$\begin{aligned} \text{9,919.21} & = \text{7,000}\left( 1 + i \right)^{\text{5}} \\ \frac{\text{9,919.21}}{\text{7,000}} & = \left( 1 + i \right)^{\text{5}} \\ 1,41703\ldots & = \left( 1 + i \right)^{\text{5}} \end{aligned}$

STEP: <no title>
[−2 points ⇒ 0 / 4 points left]

At this point we must use our exponent skills from earlier in the year! To remove the exponent on the right side of the equation, we need to take a root to kill the exponent of $\text{5}$ . (Rather than a root, we can also use an exponent of $\frac{1}{\text{5}}$ - it means the same thing.)

\begin{aligned} \sqrt[5]{1, 41703 \dots} & = \sqrt[5]{{(1 + i)}^{5}} \\ 1, 07220 \dots & = 1 + i \\ 0, 07220 \dots & = i \end{aligned}

$\begin{aligned} \sqrt[\text{5}]{1,41703\ldots} & = \sqrt[\text{5}]{\left( 1 + i \right)^{\text{5}}} \\ 1,07220\ldots & = 1 + i \\ 0,07220\ldots & = i \end{aligned}$

Finally, change the decimal into a percentage (multiply by $\text{100}$ ) and then round to two decimal places: $i = 100(0,07220\ldots) = \text{7.22}\%$ .

Submit your answer as:

ID is: 1524 Seed is: 4277

Compound interest: finding $i$

Nicky deposits $\text{N}\llap{=}\,\text{4,000}$ into a savings account which earns compound interest. The account grows to a value of $\text{N}\llap{=}\,\text{6,429.04}$ in $\text{6}$ years. Determine the value of $i$ , the rate of interest paid by the bank.

INSTRUCTION: Give your answer as a percentage rounded to two decimal places.

Answer: % per annum.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Does this account involve simple interest or compound interest? You need to know that so you know which formula to use.

STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Write down the given information.

$A = \text{N}\llap{=}\,\text{6,429.04}$
$P = \text{N}\llap{=}\,\text{4,000}$
$n = \text{6}$
$i = ?$

The question says, "earns compound interest," so we need to use the compound interest formula. Substitute in what you know!

\begin{aligned} A & = P (1 + i)^{n} \\ 6,429.04 & = 4,000 {(1 + i)}^{6} \end{aligned}

$\begin{aligned} A & = P(1+i)^{n} \\ \text{6,429.04} & = \text{4,000}\left( 1 + i \right)^{\text{6}} \end{aligned}$

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

Solve for $i$ . The first step is to divide by $\text{4,000}$ . Then we will deal with the exponent.

\begin{aligned} 6,429.04 & = 4,000 {(1 + i)}^{6} \\ \frac{6,429.04}{4,000} & = {(1 + i)}^{6} \\ 1, 60726 \dots & = {(1 + i)}^{6} \end{aligned}

$\begin{aligned} \text{6,429.04} & = \text{4,000}\left( 1 + i \right)^{\text{6}} \\ \frac{\text{6,429.04}}{\text{4,000}} & = \left( 1 + i \right)^{\text{6}} \\ 1,60726\ldots & = \left( 1 + i \right)^{\text{6}} \end{aligned}$

STEP: <no title>
[−2 points ⇒ 0 / 4 points left]

At this point we must use our exponent skills from earlier in the year! To remove the exponent on the right side of the equation, we need to take a root to kill the exponent of $\text{6}$ . (Rather than a root, we can also use an exponent of $\frac{1}{\text{6}}$ - it means the same thing.)

\begin{aligned} \sqrt[6]{1, 60726 \dots} & = \sqrt[6]{{(1 + i)}^{6}} \\ 1, 08230 \dots & = 1 + i \\ 0, 08230 \dots & = i \end{aligned}

$\begin{aligned} \sqrt[\text{6}]{1,60726\ldots} & = \sqrt[\text{6}]{\left( 1 + i \right)^{\text{6}}} \\ 1,08230\ldots & = 1 + i \\ 0,08230\ldots & = i \end{aligned}$

Finally, change the decimal into a percentage (multiply by $\text{100}$ ) and then round to two decimal places: $i = 100(0,08230\ldots) = \text{8.23}\%$ .

Submit your answer as:

ID is: 1524 Seed is: 9169

Compound interest: finding $i$

Deborah deposits $\text{N}\llap{=}\,\text{6,500}$ into a savings account which gets compound interest. The account grows to a value of $\text{N}\llap{=}\,\text{9,670.61}$ in $\text{5}$ years. Determine the interest rate the account earns.

INSTRUCTION: Give your answer as a percentage rounded to two decimal places.

Answer: % per annum.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Does this account involve simple interest or compound interest? You need to know that so you know which formula to use.

STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Write down the given information.

$A = \text{N}\llap{=}\,\text{9,670.61}$
$P = \text{N}\llap{=}\,\text{6,500}$
$n = \text{5}$
$i = ?$

The question says, "gets compound interest," so we need to use the compound interest formula. Substitute in what you know!

\begin{aligned} A & = P (1 + i)^{n} \\ 9,670.61 & = 6,500 {(1 + i)}^{5} \end{aligned}

$\begin{aligned} A & = P(1+i)^{n} \\ \text{9,670.61} & = \text{6,500}\left( 1 + i \right)^{\text{5}} \end{aligned}$

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

Solve for $i$ . The first step is to divide by $\text{6,500}$ . Then we will deal with the exponent.

\begin{aligned} 9,670.61 & = 6,500 {(1 + i)}^{5} \\ \frac{9,670.61}{6,500} & = {(1 + i)}^{5} \\ 1, 48778 \dots & = {(1 + i)}^{5} \end{aligned}

$\begin{aligned} \text{9,670.61} & = \text{6,500}\left( 1 + i \right)^{\text{5}} \\ \frac{\text{9,670.61}}{\text{6,500}} & = \left( 1 + i \right)^{\text{5}} \\ 1,48778\ldots & = \left( 1 + i \right)^{\text{5}} \end{aligned}$

STEP: <no title>
[−2 points ⇒ 0 / 4 points left]

\begin{aligned} \sqrt[5]{1, 48778 \dots} & = \sqrt[5]{{(1 + i)}^{5}} \\ 1, 08269 \dots & = 1 + i \\ 0, 08269 \dots & = i \end{aligned}

$\begin{aligned} \sqrt[\text{5}]{1,48778\ldots} & = \sqrt[\text{5}]{\left( 1 + i \right)^{\text{5}}} \\ 1,08269\ldots & = 1 + i \\ 0,08269\ldots & = i \end{aligned}$

Finally, change the decimal into a percentage (multiply by $\text{100}$ ) and then round to two decimal places: $i = 100(0,08269\ldots) = \text{8.27}\%$ .

Submit your answer as:

ID is: 1522 Seed is: 2527

Compound depreciation: population loss

The number of pelicans at the Berg river mouth is decreasing at a compound rate of $\text{15}\%$ p.a. If there are now $\text{7,000}$ pelicans, how many will there be in $\text{6}$ years' time?

Answer: There will be pelicans.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The question mentions a "compound rate" and "dereasing" population: these are both indications that the population behaves the same way as the value of a depreciating asset.

STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Determine what has been provided and what is required:

\begin{aligned} P & = 7,000 \\ i & = 15 % = 0.15 \\ n & = 6 \\ A & is required \end{aligned}

$\begin{aligned} P & = \text{7,000}\\ i & = \text{15}\% = \text{0.15}\\ n & = \text{6}\\ A &\text{ is required} \end{aligned}$

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Determine how to approach the problem:

\begin{aligned} A & = P (1 - i)^{n} \\ A & = 7,000 (1 - 0.15)^{6} \end{aligned}

$\begin{aligned} A & = P(1 - i)^n \\ A & = \text{7,000}(1 - \text{0.15})^{\text{6}} \end{aligned}$

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Solve the problem:

\begin{aligned} A & = 7,000 (0.85)^{6} \\ = 7,000 \times 0.37715 . . . ⟵_{off yet!}^{Don't round} \\ = 2,640.05 \end{aligned}

$\begin{aligned} A & = \text{7,000}( \text{0.85})^{\text{6}}\\ & = \text{7,000} \times \text{0.37715}... \quad \longleftarrow ^{\text{Don't round}}_{\text{off yet!}}\\ & = \text{2,640.05} \end{aligned}$

There will be approximately $\text{2,640}$ pelicans in $\text{6}$ years' time. (You must round to the nearest integer because you cannot have $\text{2,640.05}$ pelicans!)

Submit your answer as:

ID is: 1522 Seed is: 3356

Compound depreciation: population loss

The number of pelicans at the Sundays river mouth is decreasing at a compound rate of $\text{11}\%$ p.a. If there are now $\text{4,000}$ pelicans, how many will there be in $\text{25}$ years' time?

Answer: There will be pelicans.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The question mentions a "compound rate" and "dereasing" population: these are both indications that the population behaves the same way as the value of a depreciating asset.

STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Determine what has been provided and what is required:

\begin{aligned} P & = 4,000 \\ i & = 11 % = 0.11 \\ n & = 25 \\ A & is required \end{aligned}

$\begin{aligned} P & = \text{4,000}\\ i & = \text{11}\% = \text{0.11}\\ n & = \text{25}\\ A &\text{ is required} \end{aligned}$

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Determine how to approach the problem:

\begin{aligned} A & = P (1 - i)^{n} \\ A & = 4,000 (1 - 0.11)^{25} \end{aligned}

$\begin{aligned} A & = P(1 - i)^n \\ A & = \text{4,000}(1 - \text{0.11})^{\text{25}} \end{aligned}$

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Solve the problem:

\begin{aligned} A & = 4,000 (0.89)^{25} \\ = 4,000 \times 0.054294 . . . ⟵_{off yet!}^{Don't round} \\ = 217.18 \end{aligned}

$\begin{aligned} A & = \text{4,000}( \text{0.89})^{\text{25}}\\ & = \text{4,000} \times \text{0.054294}... \quad \longleftarrow ^{\text{Don't round}}_{\text{off yet!}}\\ & = \text{217.18} \end{aligned}$

There will be approximately $\text{217}$ pelicans in $\text{25}$ years' time. (You must round to the nearest integer because you cannot have $\text{217.18}$ pelicans!)

Submit your answer as:

ID is: 1522 Seed is: 4355

Compound depreciation: population loss

The number of cormorants at the Berg river mouth is decreasing at a compound rate of $\text{11}\%$ p.a. If there are now $\text{2,000}$ cormorants, how many will there be in $\text{7}$ years' time?

Answer: There will be cormorants.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The question mentions a "compound rate" and "dereasing" population: these are both indications that the population behaves the same way as the value of a depreciating asset.

STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Determine what has been provided and what is required:

\begin{aligned} P & = 2,000 \\ i & = 11 % = 0.11 \\ n & = 7 \\ A & is required \end{aligned}

$\begin{aligned} P & = \text{2,000}\\ i & = \text{11}\% = \text{0.11}\\ n & = \text{7}\\ A &\text{ is required} \end{aligned}$

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Determine how to approach the problem:

\begin{aligned} A & = P (1 - i)^{n} \\ A & = 2,000 (1 - 0.11)^{7} \end{aligned}

$\begin{aligned} A & = P(1 - i)^n \\ A & = \text{2,000}(1 - \text{0.11})^{\text{7}} \end{aligned}$

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Solve the problem:

\begin{aligned} A & = 2,000 (0.89)^{7} \\ = 2,000 \times 0.442313 . . . ⟵_{off yet!}^{Don't round} \\ = 884.63 \end{aligned}

$\begin{aligned} A & = \text{2,000}( \text{0.89})^{\text{7}}\\ & = \text{2,000} \times \text{0.442313}... \quad \longleftarrow ^{\text{Don't round}}_{\text{off yet!}}\\ & = \text{884.63} \end{aligned}$

There will be approximately $\text{885}$ cormorants in $\text{7}$ years' time. (You must round to the nearest integer because you cannot have $\text{884.63}$ cormorants!)

Submit your answer as:

ID is: 1535 Seed is: 5301

Simple depreciation

A delivery van is worth $\text{N}\llap{=}\,\text{80,000}$ now. It depreciates at a rate of $\text{4}\%$ p.a. on straight-line depreciation. What is the delivery van worth in $\text{22}$ years' time?

Answer: The delivery van will be worth

N =

$\text{N}\llap{=}$ .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

This is a simple depreciation question: write down the formula for simple depreciation, and look for the values you know from the question which you can substitute into the formula.

STEP: Examine the given information
[−1 point ⇒ 2 / 3 points left]

Determine what has been provided and what is required:

\begin{aligned} the value now & = 80,000 \\ rate of depreciation & = 4 % = 0.04 \\ number of years & = 22 \\ the final value & = ? \end{aligned}

$\begin{aligned} \text{the value now} & = \text{80,000}\\ \text{rate of depreciation} & = \text{4}\% = \text{0.04}\\ \text{number of years} & = \text{22}\\ \text{the final value} & = ? \end{aligned}$

STEP: Substitute in to the formula for simple depreciation
[−1 point ⇒ 1 / 3 points left]

This is a simple depreciation problem. Write down the formula and substitute in what you know.

\begin{aligned} A & = P (1 - i n) \\ A & = 80,000 (1 - (0.04 \times 22)) \end{aligned}

$\begin{aligned} A & = P(1 - in) \\ A & = \text{80,000}\left(1 - (\text{0.04} \times \text{22})\right) \end{aligned}$

STEP: Solve for

A

$A$
[−1 point ⇒ 0 / 3 points left]

Solve the problem:

\begin{aligned} A & = 80,000 (1 - 0.88) \\ = 80,000 (0.12) \\ = 9,600 \end{aligned}

$\begin{aligned} A & = \text{80,000}(1 - \text{0.88})\\ & = \text{80,000} (\text{0.12})\\ & = \text{9,600} \end{aligned}$

In $\text{22}$ years' time the delivery van will be worth $\text{N}\llap{=}\,\text{9,600}$ .

Submit your answer as:

ID is: 1535 Seed is: 2643

Simple depreciation

A delivery van is worth $\text{N}\llap{=}\,\text{11,000}$ now. It depreciates at a rate of $\text{6}\%$ p.a. on straight-line depreciation. What is the delivery van worth in $\text{15}$ years' time?

Answer: The delivery van will be worth

N =

$\text{N}\llap{=}$ .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

This is a simple depreciation question: write down the formula for simple depreciation, and look for the values you know from the question which you can substitute into the formula.

STEP: Examine the given information
[−1 point ⇒ 2 / 3 points left]

Determine what has been provided and what is required:

\begin{aligned} the value now & = 11,000 \\ rate of depreciation & = 6 % = 0.06 \\ number of years & = 15 \\ the final value & = ? \end{aligned}

$\begin{aligned} \text{the value now} & = \text{11,000}\\ \text{rate of depreciation} & = \text{6}\% = \text{0.06}\\ \text{number of years} & = \text{15}\\ \text{the final value} & = ? \end{aligned}$

STEP: Substitute in to the formula for simple depreciation
[−1 point ⇒ 1 / 3 points left]

This is a simple depreciation problem. Write down the formula and substitute in what you know.

\begin{aligned} A & = P (1 - i n) \\ A & = 11,000 (1 - (0.06 \times 15)) \end{aligned}

$\begin{aligned} A & = P(1 - in) \\ A & = \text{11,000}\left(1 - (\text{0.06} \times \text{15})\right) \end{aligned}$

STEP: Solve for

A

$A$
[−1 point ⇒ 0 / 3 points left]

Solve the problem:

\begin{aligned} A & = 11,000 (1 - 0.9) \\ = 11,000 (0.1) \\ = 1,100 \end{aligned}

$\begin{aligned} A & = \text{11,000}(1 - \text{0.9})\\ & = \text{11,000} (\text{0.1})\\ & = \text{1,100} \end{aligned}$

In $\text{15}$ years' time the delivery van will be worth $\text{N}\llap{=}\,\text{1,100}$ .

Submit your answer as:

ID is: 1535 Seed is: 9074

Simple depreciation

A truck is worth $\text{N}\llap{=}\,\text{64,000}$ now. It depreciates at a rate of $\text{12}\%$ p.a. on straight-line depreciation. What is the truck worth in $\text{2}$ years' time?

Answer: The truck will be worth

N =

$\text{N}\llap{=}$ .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

This is a simple depreciation question: write down the formula for simple depreciation, and look for the values you know from the question which you can substitute into the formula.

STEP: Examine the given information
[−1 point ⇒ 2 / 3 points left]

Determine what has been provided and what is required:

\begin{aligned} the value now & = 64,000 \\ rate of depreciation & = 12 % = 0.12 \\ number of years & = 2 \\ the final value & = ? \end{aligned}

$\begin{aligned} \text{the value now} & = \text{64,000}\\ \text{rate of depreciation} & = \text{12}\% = \text{0.12}\\ \text{number of years} & = \text{2}\\ \text{the final value} & = ? \end{aligned}$

STEP: Substitute in to the formula for simple depreciation
[−1 point ⇒ 1 / 3 points left]

This is a simple depreciation problem. Write down the formula and substitute in what you know.

\begin{aligned} A & = P (1 - i n) \\ A & = 64,000 (1 - (0.12 \times 2)) \end{aligned}

$\begin{aligned} A & = P(1 - in) \\ A & = \text{64,000}\left(1 - (\text{0.12} \times \text{2})\right) \end{aligned}$

STEP: Solve for

A

$A$
[−1 point ⇒ 0 / 3 points left]

Solve the problem:

\begin{aligned} A & = 64,000 (1 - 0.24) \\ = 64,000 (0.76) \\ = 48,640 \end{aligned}

$\begin{aligned} A & = \text{64,000}(1 - \text{0.24})\\ & = \text{64,000} (\text{0.76})\\ & = \text{48,640} \end{aligned}$

In $\text{2}$ years' time the truck will be worth $\text{N}\llap{=}\,\text{48,640}$ .

Submit your answer as:

ID is: 1536 Seed is: 4812

Compound depreciation

Farmer Roy buys a plough for $\text{N}\llap{=}\,\text{177,000}$ . The plough depreciates by $\text{20}\%$ per year following the compound depreciation method. What is the depreciated value of the plough after $\text{13}$ years?

Answer:

The depreciated value of the plough is $\text{N}\llap{=}$ .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Reread the question and identify the important words in the question. Then write down the things which you know. Then decide which equation you think is correct, and use it!

STEP: Write down the information from the question
[−1 point ⇒ 2 / 3 points left]

This is a financial maths question. So there are specific values we expect to need. To start, organise the information by writing down what we know.

\begin{aligned} original value & = N = 177,000 \\ rate of depreciation & = 20 % = 0.2 \\ number of years & = 13 \\ the final value & = ? \end{aligned}

$\begin{aligned} \text{original value} & = \text{N}\llap{=}\,\text{177,000}\\ \text{rate of depreciation} & = \text{20}\% = \text{0.2}\\ \text{number of years} & = \text{13}\\ \text{the final value} & = ? \end{aligned}$

STEP: Substitute into the depreciation formula
[−1 point ⇒ 1 / 3 points left]

This is a compound depreciation question. Write down the compound depreciation formula and go for it!

\begin{aligned} A & = P (1 - i)^{n} \\ A & = 177,000 (1 - 0.2)^{13} \end{aligned}

$\begin{aligned} A & = P(1 - i)^{n} \\ A & = \text{177,000}(1 - \text{0.2})^{\text{13}} \end{aligned}$

STEP: Evaluate to get the answer
[−1 point ⇒ 0 / 3 points left]

Solve the problem:

\begin{aligned} A & = 177,000 (0.8)^{13} \\ = 177,000 \times 0.05497... \\ = 9,730.6779... \end{aligned}

$\begin{aligned} A & = \text{177,000}( \text{0.8})^{\text{13}}\\ & = \text{177,000} \times \text{0.05497...}\\ & = \text{9,730.6779...} \end{aligned}$

Write the final answer (remember to round off to two decimal places because the answer represents money):

The value of the plough after $\text{13}$ years is $\text{N}\llap{=}\,\text{9,730.68}$ .

Submit your answer as:

ID is: 1536 Seed is: 9397

Compound depreciation

Farmer Saymore buys a combine harvester for $\text{N}\llap{=}\,\text{151,000}$ . The combine harvester depreciates by $\text{6}\%$ per year where the depreciation happens at a compound rate. What is the depreciated value of the combine harvester after $\text{13}$ years?

Answer:

The depreciated value of the combine harvester is $\text{N}\llap{=}$ .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Reread the question and identify the important words in the question. Then write down the things which you know. Then decide which equation you think is correct, and use it!

STEP: Write down the information from the question
[−1 point ⇒ 2 / 3 points left]

This is a financial maths question. So there are specific values we expect to need. To start, organise the information by writing down what we know.

\begin{aligned} original value & = N = 151,000 \\ rate of depreciation & = 6 % = 0.06 \\ number of years & = 13 \\ the final value & = ? \end{aligned}

$\begin{aligned} \text{original value} & = \text{N}\llap{=}\,\text{151,000}\\ \text{rate of depreciation} & = \text{6}\% = \text{0.06}\\ \text{number of years} & = \text{13}\\ \text{the final value} & = ? \end{aligned}$

STEP: Substitute into the depreciation formula
[−1 point ⇒ 1 / 3 points left]

This is a compound depreciation question. Write down the compound depreciation formula and go for it!

\begin{aligned} A & = P (1 - i)^{n} \\ A & = 151,000 (1 - 0.06)^{13} \end{aligned}

$\begin{aligned} A & = P(1 - i)^{n} \\ A & = \text{151,000}(1 - \text{0.06})^{\text{13}} \end{aligned}$

STEP: Evaluate to get the answer
[−1 point ⇒ 0 / 3 points left]

Solve the problem:

\begin{aligned} A & = 151,000 (0.94)^{13} \\ = 151,000 \times 0.44736... \\ = 67,552.1294... \end{aligned}

$\begin{aligned} A & = \text{151,000}( \text{0.94})^{\text{13}}\\ & = \text{151,000} \times \text{0.44736...}\\ & = \text{67,552.1294...} \end{aligned}$

Write the final answer (remember to round off to two decimal places because the answer represents money):

The value of the combine harvester after $\text{13}$ years is $\text{N}\llap{=}\,\text{67,552.13}$ .

Submit your answer as:

ID is: 1536 Seed is: 8183

Compound depreciation

Farmer Chibuzo buys a tractor for $\text{N}\llap{=}\,\text{145,000}$ . The tractor depreciates by $\text{24}\%$ per year following the compound depreciation method. What is the depreciated value of the tractor after $\text{10}$ years?

Answer:

The depreciated value of the tractor is $\text{N}\llap{=}$ .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Reread the question and identify the important words in the question. Then write down the things which you know. Then decide which equation you think is correct, and use it!

STEP: Write down the information from the question
[−1 point ⇒ 2 / 3 points left]

This is a financial maths question. So there are specific values we expect to need. To start, organise the information by writing down what we know.

\begin{aligned} original value & = N = 145,000 \\ rate of depreciation & = 24 % = 0.24 \\ number of years & = 10 \\ the final value & = ? \end{aligned}

$\begin{aligned} \text{original value} & = \text{N}\llap{=}\,\text{145,000}\\ \text{rate of depreciation} & = \text{24}\% = \text{0.24}\\ \text{number of years} & = \text{10}\\ \text{the final value} & = ? \end{aligned}$

STEP: Substitute into the depreciation formula
[−1 point ⇒ 1 / 3 points left]

This is a compound depreciation question. Write down the compound depreciation formula and go for it!

\begin{aligned} A & = P (1 - i)^{n} \\ A & = 145,000 (1 - 0.24)^{10} \end{aligned}

$\begin{aligned} A & = P(1 - i)^{n} \\ A & = \text{145,000}(1 - \text{0.24})^{\text{10}} \end{aligned}$

STEP: Evaluate to get the answer
[−1 point ⇒ 0 / 3 points left]

Solve the problem:

\begin{aligned} A & = 145,000 (0.76)^{10} \\ = 145,000 \times 0.06428... \\ = 9,321.8889... \end{aligned}

$\begin{aligned} A & = \text{145,000}( \text{0.76})^{\text{10}}\\ & = \text{145,000} \times \text{0.06428...}\\ & = \text{9,321.8889...} \end{aligned}$

Write the final answer (remember to round off to two decimal places because the answer represents money):

The value of the tractor after $\text{10}$ years is $\text{N}\llap{=}\,\text{9,321.89}$ .

Submit your answer as:

ID is: 1525 Seed is: 4137

Simple interest: finding $i$

Bongani deposits $\text{N}\llap{=}\,\text{7,000}$ into a savings account which earns simple interest. The account grows to a value of $\text{N}\llap{=}\,\text{10,192.00}$ in $\text{6}$ years. What is the interest rate for this account?

INSTRUCTION: Give your answer as a percentage rounded to one decimal place.

Answer: % per annum.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Does this account involve simple interest or compound interest? You need to know that so you know which formula to use.

STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Write down the given information.

$A = 10192$
$P = \text{7,000}$
$n = \text{6}$
$i = ?$

The question says, "earns simple interest," so we need to use the simple interest formula. Substitute in what you know!

\begin{aligned} A & = P (1 + i n) \\ 10192 & = 7,000 (1 + i (6)) \end{aligned}

$\begin{aligned} A & = P(1+in) \\ 10192 & = \text{7,000}\left( 1 + i (\text{6}) \right) \end{aligned}$

STEP: <no title>
[−2 points ⇒ 0 / 3 points left]

Solve for $i$ . The first step is to divide by $\text{7,000}$ which will let us get into the brackets. Then solve it!

\begin{aligned} 10192 & = 7,000 (1 + 6 i) \\ \frac{10192}{7,000} & = (1 + 6 i) \\ 1.456 & = 1 + 6 i \\ 0.456 & = 6 i \\ \frac{0.456}{6} & = i \\ 0.076 & = i \end{aligned}

$\begin{aligned} 10192 & = \text{7,000}\left( 1 + \text{6}i \right) \\ \frac{10192}{\text{7,000}} & = \left( 1 + \text{6}i \right) \\ \text{1.456} & = 1 +\text{6}i \\ \text{0.456} & = \text{6}i \\ \frac{\text{0.456}}{\text{6}} & = i \\ \text{0.076} & = i \end{aligned}$

Finally, multiply by $\text{100}$ to change into a percentage. Then round to one decimal place if necessary. $i = 100(\text{0.076}) = \text{7.6}\%$ .

Submit your answer as:

ID is: 1525 Seed is: 5736

Simple interest: finding $i$

Adedamola deposits $\text{N}\llap{=}\,\text{7,000}$ into a savings account which pays simple interest. The account grows to a value of $\text{N}\llap{=}\,\text{13,300.00}$ in $\text{10}$ years. Determine the interest rate the account earns.

INSTRUCTION: Give your answer as a percentage rounded to one decimal place.

Answer: % per annum.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Does this account involve simple interest or compound interest? You need to know that so you know which formula to use.

STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Write down the given information.

$A = 13300$
$P = \text{7,000}$
$n = \text{10}$
$i = ?$

The question says, "pays simple interest," so we need to use the simple interest formula. Substitute in what you know!

\begin{aligned} A & = P (1 + i n) \\ 13300 & = 7,000 (1 + i (10)) \end{aligned}

$\begin{aligned} A & = P(1+in) \\ 13300 & = \text{7,000}\left( 1 + i (\text{10}) \right) \end{aligned}$

STEP: <no title>
[−2 points ⇒ 0 / 3 points left]

Solve for $i$ . The first step is to divide by $\text{7,000}$ which will let us get into the brackets. Then solve it!

\begin{aligned} 13300 & = 7,000 (1 + 10 i) \\ \frac{13300}{7,000} & = (1 + 10 i) \\ 1.9 & = 1 + 10 i \\ 0.9 & = 10 i \\ \frac{0.9}{10} & = i \\ 0.09 & = i \end{aligned}

$\begin{aligned} 13300 & = \text{7,000}\left( 1 + \text{10}i \right) \\ \frac{13300}{\text{7,000}} & = \left( 1 + \text{10}i \right) \\ \text{1.9} & = 1 +\text{10}i \\ \text{0.9} & = \text{10}i \\ \frac{\text{0.9}}{\text{10}} & = i \\ \text{0.09} & = i \end{aligned}$

Finally, multiply by $\text{100}$ to change into a percentage. Then round to one decimal place if necessary. $i = 100(\text{0.09}) = \text{9.0}\%$ .

Submit your answer as:

ID is: 1525 Seed is: 7387

Simple interest: finding $i$

Sehlolo deposits $\text{N}\llap{=}\,\text{5,000}$ into a savings account which pays simple interest. The account grows to a value of $\text{N}\llap{=}\,\text{6,340.00}$ in $\text{4}$ years. Determine the interest rate the account earns.

INSTRUCTION: Give your answer as a percentage rounded to one decimal place.

Answer: % per annum.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Does this account involve simple interest or compound interest? You need to know that so you know which formula to use.

STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Write down the given information.

$A = 6340$
$P = \text{5,000}$
$n = \text{4}$
$i = ?$

The question says, "pays simple interest," so we need to use the simple interest formula. Substitute in what you know!

\begin{aligned} A & = P (1 + i n) \\ 6340 & = 5,000 (1 + i (4)) \end{aligned}

$\begin{aligned} A & = P(1+in) \\ 6340 & = \text{5,000}\left( 1 + i (\text{4}) \right) \end{aligned}$

STEP: <no title>
[−2 points ⇒ 0 / 3 points left]

Solve for $i$ . The first step is to divide by $\text{5,000}$ which will let us get into the brackets. Then solve it!

\begin{aligned} 6340 & = 5,000 (1 + 4 i) \\ \frac{6340}{5,000} & = (1 + 4 i) \\ 1.268 & = 1 + 4 i \\ 0.268 & = 4 i \\ \frac{0.268}{4} & = i \\ 0.067 & = i \end{aligned}

$\begin{aligned} 6340 & = \text{5,000}\left( 1 + \text{4}i \right) \\ \frac{6340}{\text{5,000}} & = \left( 1 + \text{4}i \right) \\ \text{1.268} & = 1 +\text{4}i \\ \text{0.268} & = \text{4}i \\ \frac{\text{0.268}}{\text{4}} & = i \\ \text{0.067} & = i \end{aligned}$

Finally, multiply by $\text{100}$ to change into a percentage. Then round to one decimal place if necessary. $i = 100(\text{0.067}) = \text{6.7}\%$ .

Submit your answer as:

Arithmetic of finance

1. Revision

Finding the length of an investment

Finding the length of an investment

Finding the length of an investment

Timeline question: adding or withdrawing money

Timeline question: adding or withdrawing money

Timeline question: adding or withdrawing money

Finding the length of an investment

Finding the length of an investment

Finding the length of an investment

Timelines

Timelines

Timelines

Timeline questions: following the changes

Timeline questions: following the changes

Timeline questions: following the changes

Birthday investment

Birthday investment

Birthday investment

Timeline question: interest earned

Timeline question: interest earned

Timeline question: interest earned

Timelines: following money as it comes and goes

Timelines: following money as it comes and goes

Timelines: following money as it comes and goes

Savings investment

Savings investment

Savings investment

Saving for a big purchase

Saving for a big purchase

Saving for a big purchase

Simple interest investment

Simple interest investment

Simple interest investment

Compound interest rate

Compound interest rate

Compound interest rate

Timeline questions: breaking the question down

Timeline questions: breaking the question down

Timeline questions: breaking the question down

Finding the length of an investment

Finding the length of an investment

Finding the length of an investment

Compound interest investment

Compound interest investment

Compound interest investment

Timeline questions: changing interest rates

Timeline questions: changing interest rates

Timeline questions: changing interest rates

Compound interest investment

Compound interest investment

Compound interest investment

Timelines: population growth

Timelines: population growth

Timelines: population growth

2. Depreciation

Increasing values: finding i

Increasing values: finding i

Increasing values: finding i

Simple depreciation: when an asset goes to zero value!

Simple depreciation: when an asset goes to zero value!

Simple depreciation: when an asset goes to zero value!

Compound interest: finding ii i

Compound interest: finding ii i

Compound interest: finding ii i

Compound depreciation: population loss

Compound depreciation: population loss

Compound depreciation: population loss

Simple depreciation

Simple depreciation

Simple depreciation

Compound depreciation

Compound depreciation

Compound depreciation

Simple interest: finding ii i

Simple interest: finding ii i

Simple interest: finding ii i

3. Tax

4. Annuities

Compound interest: finding $i$

Compound interest: finding $i$

Compound interest: finding $i$

Simple interest: finding $i$

Simple interest: finding $i$

Simple interest: finding $i$